• CodeTop题解

    CodeTop 第一页

    206. 反转链表 - 力扣（LeetCode） (leetcode-cn.com)

     1 class Solution {
 2 public:
 3     ListNode* reverse(ListNode* pre,ListNode* head) {
 4         if(head==nullptr)
 5             return pre;
 6         ListNode *beh = head->next;
 7         head->next = pre;
 8         pre = head;
 9         head = beh;
10         return reverse(pre,head);
11     }
12     ListNode* reverseList(ListNode* head) {
13         return reverse(nullptr,head);
14     }
15 };
    View Code

    146. LRU 缓存 - 力扣（LeetCode） (leetcode-cn.com)

    因为removeNode，修改了tail->pre的指向，这时候用tail->pre会报错

     1 class LRUCache {
 2 public:
 3     struct LinkNode{
 4         int key,val;
 5         struct LinkNode *pre;
 6         struct LinkNode *next;
 7         LinkNode():key(0),val(0),pre(nullptr),next(nullptr){}
 8         LinkNode(int key,int val):key(key),val(val),pre(nullptr),next(nullptr){}
 9     };
10     LinkNode* head;
11     LinkNode* tail;
12     unordered_map<int ,LinkNode*> map;
13     int capacity;
14     LRUCache(int _capacity):capacity(_capacity) {
15         head = new LinkNode();
16         tail = new LinkNode();
17         head->next = tail;
18         tail->pre = head;
19     }
20     
21     int get(int key) {
22         if(map.count(key)==0)
23             return -1;
24         removeNode(map[key]);
25         addToHead(map[key]);
26         return map[key]->val;
27     }
28     
29     void put(int key, int value) {
30         if(map.count(key)==0){
31             LinkNode* node = new LinkNode(key,value);
32             map[key] = node;
33             addToHead(node);
34             if(map.size()>capacity){
35                 LinkNode* delNode = tail->pre;
36                 removeNode(tail->pre);
37                 map.erase(delNode->key);//这里要使用另外定义delNode，用tail->pre报错
38                 delete delNode;//这里要使用另外定义delNode，用tail->pre报错
39             }
40         }else{
41             map[key]->val = value;
42             removeNode(map[key]);
43             addToHead(map[key]);
44         }
45     }
46     void removeNode(LinkNode* node){
47         node->pre->next = node->next;
48         node->next->pre = node->pre;
49     }
50     void addToHead(LinkNode* node){
51         node->next = head->next;
52         node->pre = head;
53         head->next = node;
54         node->next->pre = node;
55 
56     }
57 };
    View Code

    3. 无重复字符的最长子串 - 力扣（LeetCode） (leetcode-cn.com)

    二刷代码更好了，初始化方面老忘记。

     1 class Solution {
 2 public:
 3     int lengthOfLongestSubstring(string s) {
 4         if(s.size() == 0)
 5             return 0;
 6         unordered_map<char,int> map;
 7         vector<int> dp(s.size(),0);
 8         dp[0] = 1;
 9         map[s[0]] = 0;//初始化忘记了
10         int start = 0;
11         int ans = 1;//初始化1
12         for(int i=1;i<s.size();i++){
13             if(map.count(s[i]) == 0){
14                 dp[i] = dp[i-1]+1;
15             }else{
16                 start = max(start,map[s[i]]+1);
17                 dp[i] = i-start+1;
18             }
19             map[s[i]] = i;
20             ans = max(ans,dp[i]);
21         }
22         return ans;
23     }
24 };
    View Code

    215. 数组中的第K个最大元素 - 力扣（LeetCode） (leetcode-cn.com)

    这道题各种排序解法写在排序算法里

     1 class Solution {//parent = (son-1)/2    lson = parent*2+1    rson = parent*2+2
 2 public:
 3     void adjust(vector<int>&nums,int n,int index){//递归调整当前节点
 4         if(index>=n)
 5             return;
 6         int lson = index*2+1,rson = index*2+2;
 7         int mn = index;
 8         if(lson<n&&nums[lson]<nums[mn]){//堆是大还是小就看这里符号
 9             mn = lson;
10         }
11         if(rson<n&&nums[rson]<nums[mn]){
12             mn = rson;
13         }
14         if(mn!=index){
15             swap(nums[mn],nums[index]);
16             adjust(nums,n,mn);
17         }
18     }
19     void build_heap(vector<int>&nums,int n){//小根堆，排序后是从大到小
20         int last_node = n-1;
21         int last_parent = (last_node-1)/2;
22         for(int i=last_parent;i>=0;i--){//从最后一个节点的父节点（即最后一个非叶子节点进行调整）
23             adjust(nums,n,i);
24         }
25     }
26     void hsort(vector<int>&nums,int n){//在数组已经是堆的基础之上，拿到最小值赋值到数组末尾，调整堆顶，
27         build_heap(nums,n);
28         for(int i=n-1;i>=0;i--){
29             swap(nums[i],nums[0]);
30             adjust(nums,i,0);
31         }
32     }
33     int findKthLargest(vector<int>& nums, int k) {
34         int n=nums.size();
35         hsort(nums,n);
36         for(int i=0;i<nums.size();i++)
37             cout<<nums[i]<<" ";
38         return nums[k-1];
39     }
40 };
    View Code

    25. K 个一组翻转链表 - 力扣（LeetCode） (leetcode-cn.com)

     1 class Solution {
 2 public:
 3     ListNode* reverseKGroup(ListNode* head, int k) {
 4         ListNode *t  =head;
 5         for(int i=0;i<k;i++){
 6             if(t==nullptr)
 7                 return head;
 8             t=t->next;
 9         }
10         ListNode *pre = head;
11         ListNode *cur = head->next;
12         for(int i=0;i<k-1;i++){
13             
14             cur->next = pre;
15             pre = cur;
16             cur = nextcur;
17         }
18         ListNode* next = reverseKGroup(cur,k);
19         head->next = next;
20         return pre ;
21     }
22 };
    View Code

    15. 三数之和 - 力扣（LeetCode） (leetcode-cn.com)

     1 class Solution {
 2 public:
 3     vector<vector<int>> threeSum(vector<int>& nums) {
 4         vector<vector<int>> result;
 5         sort(nums.begin(),nums.end());
 6         for(int i=0;i<nums.size()&&nums[i]<=0;i++){
 7             if(i>0&&nums[i]==nums[i-1]){//对a去重
 8                 continue ;
 9             }
10             int target = -nums[i];
11             int left = i+1,right = nums.size()-1;//left和right是下标不是nums[i+1]
12             while(left<right){
13                 if(nums[left]+nums[right]==target){
14                     result.push_back(vector<int>{nums[i],nums[left],nums[right]});
15                     left++;//忘记加
16                     right--;
17                     while (right > left && nums[right] == nums[right + 1]) right--;//对b去重 忘记了
18                     while (right > left && nums[left] == nums[left - 1]) left++;//对c去重
19                 }else if(nums[left]+nums[right]<target){
20                     left++;
21                 }else if(nums[left]+nums[right]>target){
22                     right--;
23                 }
24             }
25         }
26         return result;
27     }
28 };
    View Code

    912. 排序数组 - 力扣（LeetCode） (leetcode-cn.com)

     1 class Solution {
 2 public:
 3     void qsort(vector<int>&nums,int left,int right){
 4         if(left>=right)
 5             return ;
 6         int cmp = left,j=left;
 7         int randnum = rand() % (right - left + 1) + left; // 随机选一个作为我们的cmp，思路：选好了和left交换。不要把随机数直接作为cmp
 8         swap(nums[left], nums[randnum]);
 9         for(int i=left+1;i<=right;++i){
10             if(nums[i]<nums[cmp]){//这里小于或者小于等于都行
11                 swap(nums[++j],nums[i]);
12             }
13         }
14         swap(nums[cmp],nums[j]);
15         qsort(nums,left,j-1);//这里是j
16         qsort(nums,j+1,right);
17     }
18     vector<int> sortArray(vector<int>& nums) {
19         qsort(nums,0,nums.size()-1);
20         return nums;
21     }
22 };
    View Code

    53. 最大子数组和 - 力扣（LeetCode） (leetcode-cn.com)

     1 class Solution {
 2 public:
 3     int maxSubArray(vector<int>& nums) {
 4        int ans=nums[0],sum = 0;
 5        for(int i=0;i<nums.size();i++){
 6            sum = sum+nums[i];
 7            ans = max(ans,sum);//先计入答案，再操作sum
 8            sum=max(0,sum);
 9            
10        }
11        return ans;
12     }
13 };
    View Code

    21. 合并两个有序链表 - 力扣（LeetCode） (leetcode-cn.com)

     1 class Solution {
 2 public:
 3     ListNode* mergeTwoLists(ListNode* list1, ListNode* list2) {
 4         ListNode* head = new ListNode();
 5         ListNode *t = head;
 6         while(list1&&list2){
 7             if(list1->val<=list2->val){
 8                 t->next = list1;
 9                 list1 = list1->next;
10             }
11             else{
12                 t->next = list2;
13             list2 = list2->next;
14             }
15             t=t->next;
16         }
17         while(list1){
18             t->next = list1;
19             list1 = list1->next;
20             t = t->next;
21         }
22         while(list2){
23             t->next = list2;
24             list2 = list2->next;
25             t = t->next;
26         }
27         return head->next;
28     }
29 };
    View Code

    1. 两数之和 - 力扣（LeetCode） (leetcode-cn.com)

     1 class Solution {
 2 public:
 3     vector<int> twoSum(vector<int>& nums, int target) {
 4         unordered_map<int,int> a;//key是值，value是数组下标
 5         for(int i=0;i<nums.size();i++){
 6             if(a.count(target-nums[i])){
 7                 return{i,a[target-nums[i]]};
 8             }else{
 9                 a[nums[i]]=i;//a.insert(make_pair(nums[i],i));
10             }
11         }
12         return {};
13     }
14 };
    View Code

    102. 二叉树的层序遍历 - 力扣（LeetCode） (leetcode-cn.com)

     1 class Solution {
 2 public:
 3     void pre(TreeNode* root,int depth,vector<vector<int>>&ans){
 4         if(root==NULL)
 5             return ;
 6         if(depth>=ans.size())
 7             ans.push_back(vector<int>{});
 8         ans[depth].push_back(root->val);
 9         pre(root->left,depth+1,ans);
10         pre(root->right,depth+1,ans);
11     }
12     vector<vector<int>> levelOrder(TreeNode* root) {
13         vector<vector<int>> ans;
14         pre(root, 0, ans);
15         return ans;
16     }
17 };
    View Code

    141. 环形链表 - 力扣（LeetCode） (leetcode-cn.com)

     1 class Solution {
 2 public:
 3     bool hasCycle(ListNode *head) {
 4         if(head==NULL)
 5             return false;
 6         ListNode* fast = head->next;
 7         ListNode* slow = head;
 8         while(fast!=NULL&&fast->next!=NULL){
 9             if(slow==fast)
10                 return true;
11             slow = slow->next;
12             fast = fast->next->next;
13         }
14         return false;
15     }
16 };
    View Code

    121. 买卖股票的最佳时机 - 力扣（LeetCode） (leetcode-cn.com)

     1 class Solution {
 2 public:
 3     int maxProfit(vector<int>& prices) {//dp数组的dp[i]表示那天卖出的利润，所以维护在此之前最小值就行
 4         int ans = 0;
 5         vector<int>dp(prices.size());
 6         dp[0] = prices[0];
 7         for(int i=1;i<prices.size();i++){
 8             dp[i] = min(dp[i-1],prices[i]);
 9             ans = max(ans,prices[i]-dp[i]);
10         }
11         return ans;
12     }
13 };
    View Code

    160. 相交链表 - 力扣（LeetCode） (leetcode-cn.com)

     1 class Solution {
 2 public:
 3     ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
 4         ListNode *PA = headA;
 5         ListNode *PB = headB;
 6         while(PA!=PB){
 7             if(PA)
 8                 PA = PA->next;
 9             else
10                 PA = headB;
11             if(PB)
12                 PB = PB->next;
13             else    
14                 PB = headA;
15         }
16         return PA;
17     }
18 };
    View Code

    103. 二叉树的锯齿形层序遍历 - 力扣（LeetCode） (leetcode-cn.com)

     1 class Solution {
 2 public:
 3     vector<vector<int>> zigzagLevelOrder(TreeNode* root) {
 4         if(!root)
 5             return {};
 6         vector<vector<int>> result;
 7         queue<TreeNode*>que;
 8         int depth = 1;
 9         que.push(root);
10         while(!que.empty()){
11             vector<int>path;
12             int n = que.size();
13             for(int i=0;i<n;i++){
14                 TreeNode* t = que.front();
15                 que.pop();
16                 path.push_back(t->val);
17                 if(t->left)
18                     que.push(t->left);
19                 if(t->right)
20                     que.push(t->right);
21             }
22             if(depth%2==0){
23                 for(int i=0,j=path.size()-1;i<j;i++,j--){
24                     int temp = path[i];
25                     path[i] = path[j];
26                     path[j] = temp;
27                 }
28             }
29             result.push_back(path);
30             depth++;
31         }
32         return result;
33     }
34 };
    View Code

    20. 有效的括号 - 力扣（LeetCode） (leetcode-cn.com)

     1 class Solution {
 2 public:
 3     bool isValid(string s) {
 4         unordered_map<char,char>map = {
 5             {')','('},{'}','{'},{']','['}
 6         };
 7         stack<char> st;
 8         for(int i=0;i<s.size();i++){
 9             if(s[i]==')'||s[i]=='}'||s[i]==']'){
10                 if(!st.empty()&&st.top()==map[s[i]]){
11                     st.pop();
12                 }
13                 else
14                     return false;
15             }else  
16                 st.push(s[i]);
17         }
18         return st.empty();
19     }
20 };
    View Code

    88. 合并两个有序数组 - 力扣（LeetCode） (leetcode-cn.com)

     1 class Solution {
 2 public:
 3     void merge(vector<int>& nums1, int m, vector<int>& nums2, int n) {
 4         int i=m-1,j=n-1,k=m+n-1;
 5         while(i>=0&&j>=0){
 6             if(nums1[i]>=nums2[j])
 7                 nums1[k--] = nums1[i--];
 8             else
 9                 nums1[k--] = nums2[j--];
10             
11         }
12         while(i>=0){
13             nums1[k--] = nums1[i--];
14         }
15         while(j>=0){
16             nums1[k--] = nums2[j--];
17         }
18     }
19 };
    View Code

    236. 二叉树的最近公共祖先 - 力扣（LeetCode） (leetcode-cn.com)

     1 class Solution {
 2 public:
 3     TreeNode* f(TreeNode* root, TreeNode* p, TreeNode* q){//没有父母parent这个指针，只能通过返回值给父母，nice
 4         if(!root)
 5             return NULL;
 6         if(root==p||root==q)//父子
 7             return root;
 8         root->left = f(root->left,p,q);
 9         root->right = f(root->right,p,q);
10         if(root->left&&root->right)//兄弟情况
11             return root;
12         return  root->left?root->left:root->right;
13     }
14     TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
15         return f(root,p,q);
16     }
17 };
    View Code

    33. 搜索旋转排序数组 - 力扣（LeetCode） (leetcode-cn.com)

     1 class Solution {
 2 public:
 3     int search(vector<int>& nums, int target) {
 4         int l = 0,r = nums.size()-1;
 5         while(l<=r){
 6             int mid = (l+r)/2;
 7             if(nums[mid] == target)
 8                 return mid;
 9             if(nums[mid]>=nums[0]){
10                 if(target<=nums[mid]&&target>=nums[0]){
11                     r = mid - 1;
12                 }else{
13                     l = mid+1;
14                 }
15             }else if(nums[mid]<=nums[nums.size()-1]){
16                 if(target>=nums[mid]&&target<=nums[nums.size()-1]){
17                         l = mid + 1;
18                     }else   
19                         r = mid - 1;
20             }
21         }
22         return -1;
23     }
24 };
    View Code

    5. 最长回文子串 - 力扣（LeetCode） (leetcode-cn.com)

     1 class Solution {
 2 public:
 3     string longestPalindrome(string s) {
 4         int left = 0,right = 0,len = 0;
 5         vector<vector<int>> dp(s.size(),vector<int>(s.size(),0));
 6         for(int i=0;i<s.size();i++){
 7             for(int j=i;j>=0;j--){//1234,要判断14先做了23的判断
 8                 if(s[i]==s[j]){
 9                     if(i-j<=2)
10                          dp[j][i] = 1;
11                     else if(dp[j+1][i-1]==1)
12                         dp[j][i] = 1;
13                 }
14                 if(dp[j][i]==1&&i-j+1>len){
15                     len = i-j+1;
16                     left = j;
17                     right = i;
18                 }
19             }
20         }
21         return s.substr(left,len);
22     }
23 };
    View Code

    CodeTop 第二页

     200. 岛屿数量 - 力扣（LeetCode） (leetcode-cn.com)

     1 class Solution {
 2 public:
 3     void f(vector<vector<char>>& grid,int x,int y){
 4         if(grid[x][y] == '0')
 5             return ;
 6         grid[x][y] = '0';
 7         if(y-1 >= 0 && grid[x][y-1] == '1'){
 8             f(grid,x,y-1);
 9         }
10         if(y+1 < grid[x].size() && grid[x][y+1] == '1'){
11             f(grid,x,y+1);
12         }
13         if(x-1 >= 0 && grid[x-1][y] == '1'){
14             f(grid,x-1,y);
15         }
16         if(x+1 < grid.size() && grid[x+1][y] == '1'){
17             f(grid,x+1,y);
18         }
19     }
20     int numIslands(vector<vector<char>>& grid) {
21         int ans = 0;
22         for(int i=0;i<grid.size();i++){
23             for(int j=0;j<grid[i].size();j++){
24                 if(grid[i][j] == '1'){
25                     f(grid,i,j);
26                     ans++;
27                 }
28             }
29         }
30         return ans;
31     }
32 };
    View Code

     46. 全排列 - 力扣（LeetCode） (leetcode-cn.com)

     1 class Solution {
 2 public:
 3     vector<int> path;
 4     vector<vector<int>> result;
 5     void f(vector<int>& nums,vector<int>& visit,int sum){
 6         if(sum == nums.size()){
 7             result.push_back(path);
 8             return ;
 9         }
10         for(int i=0;i<nums.size();i++){
11             if(visit[i] == 0){
12                 visit[i] = 1;
13                 path.push_back(nums[i]);
14                 f(nums,visit,sum+1);
15                 visit[i] = 0;
16                 path.pop_back();
17             }
18         }
19     }
20     vector<vector<int>> permute(vector<int>& nums) {
21         vector<int>visit(nums.size(),0);
22         f(nums,visit,0);
23         return result;
24     }
25 };
    View Code

     415. 字符串相加 - 力扣（LeetCode） (leetcode-cn.com)

     1 class Solution {
 2 public:
 3     string addStrings(string num1, string num2) {//还有进位哈哈。。还有连续进位9999+1
 4         if(num1.size()<num2.size()){
 5             swap(num1,num2);
 6         }
 7         int n1 = num1.size()-1,n2 = num2.size()-1;
 8         while(n2>=0){
 9             num1[n1] = num1[n1]-'0'+num2[n2];
10             if(num1[n1]>'9'){
11                 num1[n1] -= 10;
12                 if(n1-1>=0)
13                     num1[n1-1] +=1;
14                 else
15                     num1 = '1' + num1;
16             }
17             n1--;n2--;
18         }
19         while(n1 >=0 &&num1[n1]>'9'){
20             num1[n1] -= 10;
21             if(n1-1>=0)
22                 num1[n1-1] +=1;
23             else
24                 num1 = '1' + num1;
25             n1--;n2--;
26         }
27         return num1;
28     }
29 };
    View Code

     92. 反转链表 II - 力扣（LeetCode） (leetcode-cn.com)

     1 class Solution {
 2 public:
 3     ListNode* reverseBetween(ListNode* head, int left, int right) {
 4         if(head==nullptr||head->next==nullptr)
 5             return head ;
 6         int cnt = right - left + 1;
 7         if(left==1){
 8             ListNode *pre = nullptr;
 9             ListNode *cur = head;
10             while(cnt--){
11                 ListNode *beh = cur->next;
12                 cur->next = pre;
13                 pre = cur;
14                 cur = beh;
15             }
16             head->next = cur;
17             return pre;
18         }
19         ListNode *head2 = head;//left == 2
20         while(left>2){//left > 2
21             head2 = head2->next;
22             left--;
23         }
24         ListNode *pre = head2;
25         ListNode *cur = head2->next;
26         while(cnt--){
27             ListNode *beh = cur->next;
28             cur->next = pre;
29             pre = cur;
30             cur = beh;
31         }
32         head2->next->next = cur;
33         head2->next = pre;
34         return head;
35     }
36 };
    View Code

     142. 环形链表 II - 力扣（LeetCode） (leetcode-cn.com)

     1 class Solution {
 2 public:
 3     ListNode *detectCycle(ListNode *head) {
 4         if(head==NULL||head->next==NULL)
 5             return NULL;
 6         ListNode *slow = head->next;
 7         ListNode *fast = head->next->next;
 8         while(slow!=fast){
 9             if(fast==NULL||fast->next==NULL)
10                 return NULL;
11             slow = slow->next;
12             fast = fast->next->next;
13         }
14         while(slow != head){
15             head = head->next;
16             slow = slow->next;
17         }
18         return head;
19     }
20 };
    View Code

     23. 合并K个升序链表 - 力扣（LeetCode） (leetcode-cn.com)

     1 /**
 2  * Definition for singly-linked list.
 3  * struct ListNode {
 4  *     int val;
 5  *     ListNode *next;
 6  *     ListNode() : val(0), next(nullptr) {}
 7  *     ListNode(int x) : val(x), next(nullptr) {}
 8  *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 9  * };
10  */
11 class Solution {
12 public:
13     ListNode* mergeKLists(vector<ListNode*>& lists) {
14         if(lists.size()==0)
15             return nullptr;
16         for(int i=0;i<lists.size()-1;i++){
17             ListNode *head = new ListNode();
18             ListNode *t = head;
19             ListNode *l1 = lists[i];
20             ListNode *l2 = lists[i+1];
21             
22             while(l1&&l2){
23                 if(l1->val<=l2->val){
24                     t->next = l1;
25                     l1 = l1->next;
26                     t = t->next;
27                 }else{
28                     t->next = l2;
29                     l2 = l2->next;
30                     t = t->next;
31                 }
32             }
33             while(l1){
34                     t->next = l1;
35                     l1 = l1->next;
36                     t = t->next;
37             }
38             while(l2){
39                     t->next = l2;
40                     l2 = l2->next;
41                     t = t->next;
42                 }
43             lists[i+1] = head->next;
44         }
45         return lists[lists.size()-1];
46     }
47 };
    View Code

    分治算法更好

    54. 螺旋矩阵 - 力扣（LeetCode） (leetcode-cn.com)

     1 class Solution {
 2 public:
 3     vector<int> ans;
 4     void f(vector<vector<int>>& matrix,int row,int col,int x,int y){//一行或者一列
 5         if(x>row||y>col){
 6             return ;
 7         }        
 8         for(int j=y;j<=col;j++){
 9             ans.push_back(matrix[x][j]);
10         }
11         x++;
12         for(int i = x;i<=row;i++){
13             ans.push_back(matrix[i][col]);
14         }
15         col--;
16          if(x>row||y>col){
17             return ;
18         }   
19         for(int j = col;j>=y;j--){
20             ans.push_back(matrix[row][j]);
21         }
22         row--;
23         for(int i = row;i>=x;i--){
24             ans.push_back(matrix[i][y]);
25         }
26         y++;
27         f(matrix,row,col,x,y);
28             
29     }
30     vector<int> spiralOrder(vector<vector<int>>& matrix) {
31         if(matrix.size()==0)
32             return ans;  
33         f(matrix,matrix.size()-1,matrix[0].size()-1,0,0);
34         return ans;
35     }
36 };
    View Code

     300. 最长递增子序列 - 力扣（LeetCode） (leetcode-cn.com)

     1 class Solution {
 2 public:
 3     int lengthOfLIS(vector<int>& nums) {
 4         int ans = 1;
 5         vector<int> dp(nums.size(),1);
 6         for(int i=0;i<nums.size();i++){
 7             for(int j=0;j<i;j++){
 8                 if(nums[i]>nums[j]){
 9                     dp[i] = max(dp[i],dp[j]+1);
10                     ans = max(ans,dp[i]);
11                 }
12             }
13         }
14         return ans;
15     }
16 };
    View Code

     42. 接雨水 - 力扣（LeetCode） (leetcode-cn.com)

     1 class Solution {
 2 public:
 3     int trap(vector<int>& height) {
 4         vector<vector<int>> dp(height.size(),vector<int>(2,0));
 5         dp[0][0] = height[0];
 6         for(int i=1;i<height.size();i++){
 7             dp[i][0] = max(dp[i-1][0],height[i]);
 8         }
 9         dp[height.size()-1][1] = height[height.size()-1];
10         for(int i=height.size()-2;i>=0;i--){
11             dp[i][1] = max(dp[i+1][1],height[i]);
12         }
13         int ans = 0;
14         for(int i=0;i<height.size();i++){
15             ans += max(min(dp[i][0],dp[i][1])-height[i],0);
16         }
17         return ans;
18     }
19 };
    View Code

     704. 二分查找 - 力扣（LeetCode） (leetcode-cn.com)

     1 class Solution {
 2 public:
 3     int search(vector<int>& nums, int target) {
 4         int n = nums.size();
 5         int l = 0,r = n-1;
 6         while(l<=r){
 7             int mid = (l+r)/2;
 8             if(target == nums[mid])
 9                 return mid;
10             if(target < nums[mid])
11                 r = mid - 1;
12             else
13                 l = mid + 1;
14         }
15         return -1;
16     }
17 };
    View Code

     94. 二叉树的中序遍历 - 力扣（LeetCode） (leetcode-cn.com)

     1 class Solution {
 2 public:
 3     void f(TreeNode* root,vector<int> &ans){
 4         if(root == nullptr)
 5             return ;
 6         f(root->left,ans);
 7         ans.push_back(root->val);
 8         f(root->right,ans);
 9     }
10     vector<int> inorderTraversal(TreeNode* root) {
11         vector<int> ans;
12         f(root,ans);
13         return ans;
14     }
15 };
    View Code

     232. 用栈实现队列 - 力扣（LeetCode） (leetcode-cn.com)

     1 class MyQueue {
 2 public:
 3     MyQueue() {
 4         
 5     }
 6     stack<int> st1;
 7     stack<int> st2;
 8     void push(int x) {
 9         st1.push(x);
10     }
11     
12     int pop() {
13         int x;
14         if(!st2.empty()){
15             x = st2.top();
16             st2.pop();
17         }else{
18             while(!st1.empty()){
19                 st2.push(st1.top());
20                 st1.pop();
21             }
22             x = st2.top();
23             st2.pop();
24         }
25         return x;
26     }
27     
28     int peek() {
29         if(!st2.empty()){
30             return st2.top();
31         }else{
32             while(!st1.empty()){
33                 st2.push(st1.top());
34                 st1.pop();
35             }
36         }
37         return st2.top();
38     }
39     
40     bool empty() {
41         if(st2.empty()&&st1.empty())
42             return true;
43         return false;
44     }
45 };
    View Code

     143. 重排链表 - 力扣（LeetCode） (leetcode-cn.com)

     1 class Solution {
 2 public:
 3     void reorderList(ListNode* head) {//先反转后部分然后再插入进前部分
 4         if(head==nullptr||head->next==nullptr)
 5             return ;
 6         ListNode* slow = head;
 7         ListNode* fast = head;
 8         ListNode* end = nullptr;//end = 2无论奇偶,前部分末尾节点
 9         while(fast!=nullptr&&fast->next!=nullptr){
10             end = slow;
11             slow = slow->next;
12             fast = fast->next->next;
13         }
14         end->next = nullptr;
15         ListNode* pre = nullptr;
16         ListNode* cur = slow;
17         while(cur){
18             ListNode *t = cur->next;
19             cur->next = pre;
20             pre = cur;
21             cur = t;
22         }
23         ListNode *h = head;
24         while(h!=end){
25             ListNode *t = pre->next;
26             pre->next = h->next;//
27             h->next = pre;
28             h = h->next->next;
29             pre = t;
30         }
31         h->next = pre;
32     }
33 };
    View Code

     199. 二叉树的右视图 - 力扣（LeetCode） (leetcode-cn.com)

     1 class Solution {
 2 public:
 3     vector<int> ans;
 4     void f(TreeNode* root,int depth){
 5         if(root == nullptr)
 6             return ;
 7         if(depth>=ans.size())
 8             ans.push_back(root->val);
 9         f(root->right,depth+1);
10         f(root->left,depth+1);
11     }
12     vector<int> rightSideView(TreeNode* root) {
13         f(root,0);
14         return ans;
15     }
16 };
    View Code

     124. 二叉树中的最大路径和 - 力扣（LeetCode） (leetcode-cn.com)

     1 class Solution {//不一定经过根节点，那就对每个节点都看一下是不是最大答案咯
 2 public://要维护两个东西，一个是给上层用的root->val，一个是给最终答案
 3 //递归，不熟练，两种情况？对于当前节点就两种情况，一种是最终答案，一种给上层用。如果当前节点和右节点都是负数，左节点正数，这种情况的答案在遍历左子树时以左节点为当前节点那会已经做了！
 4     int result = -1e8;
 5     int maxPathSum(TreeNode* root) {
 6         if(root == nullptr)
 7             return 0;
 8         int left = 0,right = 0;
 9         if(root->left){
10             maxPathSum(root->left);
11             left = max(root->left->val,0);
12         }
13         if(root->right){
14             maxPathSum(root->right);
15             right = max(root->right->val,0);
16         }
17         result = max(result,left+right+root->val);
18         root->val = max(left,right)+root->val;
19         return result;
20     }
21 };
    View Code

    70. 爬楼梯 - 力扣（LeetCode） (leetcode-cn.com)

     1 class Solution {
 2 public:
 3     int climbStairs(int n) {
 4         if(n<=2)
 5             return n;
 6         vector<int> dp(n);
 7         dp[1] = 1;dp[2] = 2;
 8         for(int i=3;i<n;i++){
 9             dp[i] = dp[i-1]+dp[i-2];
10         }
11         return dp[n-1]+dp[n-2];
12     }
13 };
    View Code

    56. 合并区间 - 力扣（LeetCode） (leetcode-cn.com)

     1 class Solution {
 2 public:
 3     static bool cmp(vector<int>&a,vector<int>&b){
 4         if(a[0]!=b[0])
 5             return a[0]<b[0];
 6         else   
 7             return a[1]<b[1];
 8     }
 9     vector<vector<int>> merge(vector<vector<int>>& intervals) {
10         sort(intervals.begin(),intervals.end(),cmp);
11         vector<vector<int>> result;
12         vector<int> path;
13         for(int i=1;i<intervals.size();i++){//1,4    2,3
14             if(intervals[i-1][1]>=intervals[i][0]){
15                 intervals[i][0] = intervals[i-1][0];
16                 intervals[i][1] = max(intervals[i][1],intervals[i-1][1]);
17             }else{
18                 result.push_back({intervals[i-1][0],intervals[i-1][1]});
19             }
20         }
21         result.push_back(vector<int>{intervals[intervals.size()-1][0],intervals[intervals.size()-1][1]});
22         return result;
23     }
24 };
    View Code

    剑指 Offer 22. 链表中倒数第k个节点 - 力扣（LeetCode） (leetcode-cn.com)

     1 class Solution {
 2 public:
 3     ListNode* getKthFromEnd(ListNode* head, int k) {
 4         ListNode *ans = head,*tail = head;
 5         while(k--){
 6             tail = tail->next;
 7         }
 8         while(tail){
 9             tail = tail->next;
10             ans = ans->next;
11         }
12         return ans;
13     }
14 };
    View Code

     19. 删除链表的倒数第 N 个结点 - 力扣（LeetCode） (leetcode-cn.com)

     1 class Solution {
 2 public:
 3     ListNode* removeNthFromEnd(ListNode* head, int n) {
 4         ListNode *t = head,*del = head,*pre = nullptr;
 5         while(n--){
 6             t = t->next;
 7         }
 8         while(t){
 9             t = t->next;
10             pre = del;
11             del = del->next;
12         }
13         if(pre)//删除头节点pre是空
14             pre->next = del->next;
15         else
16             head = head->next;
17         return head;
18     }
19 };
    View Code

     69. x 的平方根 - 力扣（LeetCode） (leetcode-cn.com)

     1 class Solution {
 2 public:
 3     int mySqrt(int x) {//二分法+用x/i<i而不是i*i>x（溢出）
 4         if(x<=1)
 5             return x;
 6         int l = 0,r = x;
 7         while(r-l>=2){//作差为1也就是只有两个数时返回小的那个
 8             int mid = (l+r)/2;
 9             if(mid == x/mid)
10                 return mid;
11             else if(mid > x/mid)
12                 r = mid;
13             else
14                 l = mid;
15         }
16         return l;
17     }
18 };
    View Code

     CodeTop第三页

     8. 字符串转换整数 (atoi) - 力扣（LeetCode） (leetcode-cn.com)

    这道题跟x的平方根有相同地方，超过int了，可以用i>x/i这种取巧方法避免越界，我偷懒用longlong了哈哈

     1 class Solution {
 2 public:
 3     int myAtoi(string s) {//2147483647    -2147483648
 4         int n = s.size();
 5         long long ans = 0,f = 1;
 6         int i = 0;
 7         while(i < n&&s[i] == ' ')
 8             i++;
 9         if(s[i] == '-'){
10             f = -1;
11             i++;
12         }  
13         else if(s[i] == '+'){
14             i++;
15         }
16         for(;i<n;i++){
17             if(s[i] >= '0'&&s[i] <= '9'){
18                 ans = ans*10 + s[i] - '0';
19                 if(ans>2147483647&&f == 1){
20                     ans = 2147483647;
21                     return ans;
22                 }
23                 if(ans>2147483648&&f == -1){
24                     ans = -2147483648;
25                     return ans;
26                 }
27                     
28             }else
29                 break;
30         }
31         ans*=f;
32         return (int)ans;
33     }
34 };
    View Code

    82. 删除排序链表中的重复元素 II - 力扣（LeetCode） (leetcode-cn.com)

     1 class Solution {
 2 public:
 3     ListNode* deleteDuplicates(ListNode* head) {
 4         ListNode *cur = head;
 5         ListNode *pre = new ListNode();
 6         pre->next = cur;
 7         head = pre;
 8         while(cur&&cur->next){
 9             if(cur->val == cur->next->val){
10                 while(cur&&cur->next&&cur->val == cur->next->val){
11                     cur = cur->next;
12                 }
13                 pre->next = cur->next;//pre不动
14                 cur = cur->next;
15             }
16             else{
17                 pre = pre->next;//pre移动
18                 cur = cur->next;
19             }
20         }
21         return head->next;
22     }
23 };
    View Code

     148. 排序链表 - 力扣（LeetCode） (leetcode-cn.com)

     1 class Solution {
 2 public:
 3     ListNode* merge(ListNode* head){
 4         if(head == nullptr||head->next == nullptr)//保证两个节点
 5             return head;
 6         ListNode *fast = head->next->next,*slow = head->next;
 7         ListNode* pre = head;
 8         while(fast!=nullptr&&fast->next!=nullptr){
 9             fast = fast->next->next;
10             pre = slow;
11             slow = slow->next;
12         }
13         pre->next = nullptr;
14         head = merge(head);
15         slow = merge(slow);
16         ListNode *h = new ListNode();
17         ListNode* ans = h;
18         while(head&&slow){
19             if(head->val < slow->val){
20                 h->next = head;
21                 head = head->next;
22                 h = h->next;
23             }else{
24                 h->next = slow;
25                 slow = slow->next;
26                 h = h->next;
27             }
28         }
29         if(head){
30             h->next = head;
31         }
32         if(slow){
33             h->next = slow;
34         }
35         return ans->next;
36     }
37     ListNode* sortList(ListNode* head) {
38         return merge(head);
39     }
40 };
    View Code

     2. 两数相加 - 力扣（LeetCode） (leetcode-cn.com)

    链表的题定义假头！

     1 class Solution {//插入节点要找到前驱节点
 2 public:
 3     ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {//假头！
 4         ListNode* ans = new ListNode();
 5         ListNode* head = ans;
 6         int y = 0;
 7         while(l1||l2){
 8             int x1=l1?l1->val:0;
 9             int x2 = l2?l2->val:0;
10             y=y+x1+x2;
11             ListNode *t = new ListNode(y%10);
12             head->next = t;
13             head = head->next;
14             l1 = l1?l1->next:nullptr;
15             l2 = l2?l2->next:nullptr;
16             y/=10;
17         }
18 
19         if(y==1){
20             ListNode *t = new ListNode(y);
21             head->next = t;
22         }
23         return ans->next;
24     }
25 };
    View Code

     72. 编辑距离 - 力扣（LeetCode） (leetcode-cn.com)

    和链表类似，搞一个空的出来。这道题搞一个空串a和b比，和空串b和a比，这样初始化。然后两层for里，i，j从1开始，if里面word1[i-1]==word2[j-1]

     1 class Solution {
 2 public:
 3     int minDistance(string word1, string word2) {
 4         if(word1.size() == 0)
 5             return word2.size();
 6         if(word2.size() == 0)
 7             return word1.size();
 8         int n1 = word1.size(),n2 = word2.size();
 9         vector<vector<int>> dp(n1+5,vector<int>(n2+5,0));
10         for(int i=1;i<=n1;i++){//初始化方面拉跨了
11                 dp[i][0] = i;
12         }
13         for(int j=1;j<=n2;j++){
14                 dp[0][j] = j;
15         }
16         for(int i = 1;i <= n1;i++){
17             for(int j = 1;j <= n2;j++){
18                 if(word1[i-1]!=word2[j-1]){
19                     dp[i][j] = min(dp[i-1][j-1],min(dp[i-1][j],dp[i][j-1])) + 1;//1替换，2删除，3增加
20                 }
21                 else{
22                     dp[i][j] = min(dp[i-1][j-1],min(dp[i-1][j]+1,dp[i][j-1]+1));
23                 }
24             }
25         }
26         return dp[n1][n2];
27     }
28 };
    View Code

     144. 二叉树的前序遍历 - 力扣（LeetCode） (leetcode-cn.com)

     1 class Solution {
 2 public:
 3     vector<int> ans;
 4     vector<int> preorderTraversal(TreeNode* root) {
 5         if(root == nullptr)
 6             return ans;
 7         ans.push_back(root->val);
 8         preorderTraversal(root->left);
 9         preorderTraversal(root->right);
10         return ans;
11     }
12 };
    View Code

     迭代法（栈实现）

     1 class Solution {
 2 public:
 3     stack<TreeNode*> st;
 4     vector<int> preorderTraversal(TreeNode* root) {
 5         vector<int> ans;
 6         if(root == nullptr)
 7             return ans;
 8         st.push(root);
 9         while(!st.empty()){
10             TreeNode* t = st.top();
11             st.pop();
12             ans.push_back(t->val);
13             if(t->right)
14                 st.push(t->right);
15             if(t->left)
16                 st.push(t->left);
17 
18         }
19         return ans;
20     }
21 };
    View Code

     1143. 最长公共子序列 - 力扣（LeetCode） (leetcode-cn.com)

     1 class Solution {
 2 public:
 3     int longestCommonSubsequence(string text1, string text2) {
 4         int ans = 0;
 5         int n1 = text1.size(),n2 = text2.size();
 6         vector<vector<int>> dp(n1+1,vector<int>(n2+1,0));
 7         for(int i=1;i<=n1;i++){
 8             for(int j=1;j<=n2;j++){
 9                 if(text1[i-1]==text2[j-1]){
10                     dp[i][j] = dp[i-1][j-1]+1;
11                 }else{
12                     dp[i][j] = max(dp[i-1][j-1],max(dp[i-1][j],dp[i][j-1]));
13                 }
14                 ans = max(ans,dp[i][j]);
15             }
16         }
17         return ans;
18     }
19 };
    View Code

     124. 二叉树中的最大路径和 - 力扣（LeetCode） (leetcode-cn.com)

    因为这里答案和递归函数返回值不一样，其他人都选择了再声明一个函数，我这边就一个函数了
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  • 原文地址：https://www.cnblogs.com/hcl6/p/15947204.html
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