LeetCode链表解题模板

一、通用方法以及题目分类

0、遍历链表

方法代码如下，head可以为空：

ListNode* p = head;
while(p!=NULL)
    p = p->next;

可以在这个代码上进行修改，比如要计算链表的长度：

ListNode* p = head;
int num = 0;
while(p!=NULL){
    num++;
    p = p->next;
}

如果要找到最后的节点，可以更改while循环中的条件，只不过需要加上head为NULL时的判断

if(!head) return head;
ListNode* p = head;
while(p->next!=NULL){
    p = p->next;
}

还可以使用两个指针，一个用来遍历，一个用来记录前一个

ListNode* p_before = NULL;
ListNode* p = head;
while(p!=NULL){
    p_before = p;
    p = p->next;
}

1、在head之前加上一个节点来简化计算

头结点的操作不同于中间节点，所以需要额外判断。穿插在很多题目中。

LeetCode 203 删除节点中的元素：https://leetcode.com/problems/remove-linked-list-elements/description/

2、两链表合并操作

l1代表链表1的头，l2代表链表2的头，通过以下方式来遍历链表（或是将while循环中的&&改成||）

while循环中如果是&&，则退出循环需要处理一个为空、一个还有的情况。如果while循环中为||，则在while循环体中需要判断其中一个为空的情况

还可以联合前一种方式，在head之前加上一个节点来简化计算

while(l1!=NULL && l2!=NULL){
    if(l1 != NULL){
        //operator on l1
        l1 = l1->next;
    } 
    if(l2 != NULL){
        //operator on l2
        l2 = l2->next;
    }
}
//operator on the remain

比如LeetCode2的两链表数相加的题：

ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
    ListNode* result = new ListNode(0);
    ListNode* current = result;
    int add = 0;
    while(l1!=NULL || l2!=NULL || add!=0){
        int num1 = 0;
        int num2 = 0;
        if(l1 != NULL){
            num1 = l1->val;
            l1 = l1->next;
        } 
        if(l2 != NULL){
            num2 = l2->val;
            l2 = l2->next;
        }
        ListNode* t = new ListNode((num1+num2+add)%10);
        add = (num1+num2+add)>=10?1:0;
        current->next = t;
        current = t;
    }
    return result->next;
}                            

LeetCode2 两数相加：https://leetcode.com/problems/add-two-numbers/description/

LeetCode21 合并两链表：https://leetcode.com/problems/merge-two-sorted-lists/description/

LeetCode86 划分链表：https://leetcode.com/problems/partition-list/description/

LeetCode328 奇偶链表：https://leetcode.com/problems/odd-even-linked-list/description/

3、前后指针

l1在前，l2在后，让l2先走n步，当退出第二个循环时，l2位空，l1为倒数第n个节点。可以判断链表是否有环，确定环出现的位置等

ListNode* l1 = head;
ListNode* l2 = head;
while(n-->0){
    l2 = l2->next;
}
while(l2 !=NULL){
    l1 = l1->next;
    l2 = l2->next;
}

利用快慢指针可以找到链表的中间节点，l1每次向前走一格，l2每次向前走两个。

当l1、l2都从head开始时，while循环退出时，如果是链表节点数是奇数，则l1指向正中间，如果是偶数，则l1指向位置为中间后一个，例：如果有4个，则l1指向第3个

ListNode* l1 = head;
ListNode* l2 = head;
while (l2 != NULL && l2->next!= NULL) {
    l1 = l1->next;
    l2 = l2->next->next;
}  

当l1从head，l2从head->next开始，while循环退出时，如果是链表节点数是奇数，则l1指向正中间，如果是偶数，则l1指向位置为正中间，例：如果有4个，则l1指向第2个

ListNode* l1 = head;
ListNode* l2 = head->next;
while (l2 != NULL && l2->next!= NULL) {
　　l1 = l1->next;
　　l2 = l2->next->next;
}

LeetCode19 移除倒数第n个数：https://leetcode.com/problems/remove-nth-node-from-end-of-list/description/

LeetCode109 将链表转化为二叉树：https://leetcode.com/problems/convert-sorted-list-to-binary-search-tree/description/

LeetCode141 判断链表是否有环：https://leetcode.com/problems/linked-list-cycle/description/

LeetCode142 找出链表有环的位置：https://leetcode.com/problems/linked-list-cycle-ii/description/

LeetCode160 找出两链表的交叉点：https://leetcode.com/problems/intersection-of-two-linked-lists/description/

287. Find the Duplicate Number https://leetcode.com/problems/find-the-duplicate-number/description/

数组的快慢指针

class Solution {
public:
    int findDuplicate(vector<int>& nums) {
        if(nums.size()<2) return -1;
        int pos1 = nums[0];
        int pos2 = nums[pos1];
        while(pos1!=pos2){
            pos1 = nums[pos1];
            pos2 = nums[nums[pos2]];
        }
        pos1 = 0;
        while(pos1!=pos2){
            pos1 = nums[pos1];
            pos2 = nums[pos2];
        }
        return pos1;
    }
};

1、数组也会形成环，同160的思路

4、利用前后指针处理链表内部数据，综合前几种处理方式

同样是l1和l2指针，一个在前一个在后，只是通过这样遍历一遍需要对链表中的数据进行修改。

思考方式：

1、假设进行中间某一点，看需要保存那些位置的值，即考虑n到n+1的情况

2、查看边界条件，开始位置的边界，结束位置的边界

反转链表的操作如下，注意将l1初始化为NULL，这样在反转之后可以让尾部节点为空

ListNode* reverse(ListNode* head){
　　if(head == NULL || head->next == NULL) return head;
　　ListNode* l1 = NULL;
　　ListNode* l2 = head;
　　while(l2){
　　　　ListNode* t = l2->next;
　　　　l2->next = l1;
　　　　l1 = l2;
　　　　l2 = t;
　　}
　　return l1;
}

 LeetCode24 反转链表中的元素：https://leetcode.com/problems/swap-nodes-in-pairs/description/

 LeetCode61 旋转链表：https://leetcode.com/problems/rotate-list/description/

LeetCode83 https://leetcode.com/problems/remove-duplicates-from-sorted-list/description/

LeetCode82 https://leetcode.com/problems/remove-duplicates-from-sorted-list-ii/description/

LeetCode206 颠倒链表：https://leetcode.com/problems/reverse-linked-list/description/

1、取中间某一个位置，如果需要颠倒链表，需要保存两个值，一个是前面的指针，一个是当前指针，让当前指针的next指向前一个指针，然后两个指针整体向后移动一位，此时由于当前的next被修改了，所以next的位置需要提前被保存起来

2、考虑边界值，开始时前一个指针需要被设置为NULL，结束时，当前指针为NULL，前一个指针指向最后一个元素。

LeetCode143 重新整理链表 https://leetcode.com/problems/reorder-list/description/

是以上几种题型的综合，既有快慢指针，又有反转链表，还有合并链表

LeetCode234 回文链表 https://leetcode.com/problems/palindrome-linked-list/description/

LeetCode445 反向相加两数 https://leetcode.com/problems/add-two-numbers-ii/description/

5、链表排序

LeetCode 147 链表的插入排序：https://leetcode.com/problems/insertion-sort-list/description/

LeetCode 148 链表的归并排序：https://leetcode.com/problems/sort-list/description/

6、其他

LeetCode 237 删除节点：https://leetcode.com/problems/delete-node-in-a-linked-list/description/

LeetCode 430 铺展多级链表：https://leetcode.com/problems/flatten-a-multilevel-doubly-linked-list/description/

二、答案以及边界情况

LeetCode2：

ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
    ListNode* result = new ListNode(0);
        ListNode* current = result;
    int add = 0;
    while(l1!=NULL || l2!=NULL || add!=0){
        int num1 = 0;
        int num2 = 0;
        if(l1 != NULL){
            num1 = l1->val;
            l1 = l1->next;
        } 
        if(l2 != NULL){
            num2 = l2->val;
            l2 = l2->next;
        }
        ListNode* t = new ListNode((num1+num2+add)%10);
        add = (num1+num2+add)>=10?1:0;
        current->next = t;
        current = t;
           }
    return result->next;
}

1、while循环中未加入add

2、add的再次计算以及新建ListNode的取值

3、加入头结点来简化不断增加链表的操作，不加时则需要额外判断第一个节点加入的情况

LeetCode19：

View Code

1、当链表长度为n时会出现删除头节点的情况，加上一个头节点可以简化这种操作

LeetCode21：

ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
        ListNode l(0);
        ListNode* current = &l;
        while(l1!=NULL&&l2!=NULL){
            if(l1->val < l2->val){
                current->next = l1;
                l1 = l1->next;
            }else{
                current->next = l2;
                l2 = l2->next;
            }
            current = current->next;
        }
        if(l1 == NULL)
            current->next = l2;
        if(l2 == NULL)
            current->next = l1;
        return l.next;
}

1、while循环中入股哦是&&，则退出循环需要处理一个为空、一个还有的情况。如果while循环中为||，则在while循环体中需要判断其中一个为空的情况

LeetCode24：

ListNode* swapPairs(ListNode* head) {
        if(head == NULL) return head;
        ListNode l1(0);
        l1.next = head;
        ListNode* before = &l1;
        ListNode* t1 = head;
        ListNode* t2 = head->next;
        while(t1!=NULL && t2!=NULL){
            t1->next = t2->next;
            t2->next = t1;
            before->next = t2;
            t1 = t1->next;
            if(t1!=NULL) t2 = t1->next;
            before = before->next->next;
        }
        return l1.next;
    }

1、明白反转链表需要哪几个节点的值再思考

2、这里需要保存before，但是出现了头节点可能为空的情况

LeetCode61：旋转链表

总体思路还是前后指针，但k可能大于链表长度，所以前指针先走的步数不确定

解法一：前进k步，如果到达链表末尾时还未走k步，则说明k大于链表长度。需要重新从头走k%lenth步。然后前后指针同时向后走，到达末尾后反转。

ListNode* rotateRight(ListNode* head, int k) {
        if(head == NULL || k == 0) return head;
        int temp = k;
        ListNode t(0);
        t.next = head;
        ListNode* p1 = head;
        ListNode* before1 = &t;
        int num = 0;
        while(p1!=NULL && k-->0){
            num++;
            before1 = before1->next;
            p1 = p1->next;
        }
        if(p1 == NULL){
            k = temp%num;
            if(temp%num == 0) return head;
            p1 = head;
            before1 = &t;
            while(k-->0){
                before1 = before1->next;
                p1 = p1->next;
            }
        }
        
        ListNode* p2 = head;
        ListNode* before2 = &t;
        while(p1!=NULL){
            p1 = p1->next;
            p2 = p2->next;
            before1 = before1->next;
            before2 = before2->next;
        }
        before2->next = NULL;
        before1->next = head;
        return p2;
    }

1、如果k==0，也就是最后一步的链表断裂和重新连接的步骤的位置都是同一个，则会形成环，这里需要特别注意（head是绝对位置，before1和before2都是相对位置，当出现这类型赋值时一定要考虑边界条件）

2、head == NULL时，num为0，在除的时候num不能为0，所以开头需要除去head==NULL的情况

3、在判断k==0时候要注意位置，在对k取余数的时候也会出现余数为0的情况，所以也需要额外的判断k==0

4、k在循环的时候是不断递减的，需要先保存以下k的值。

解法二：将链表变成循环链表。相比上一个方法，边界条件少了很多，代码也不容易写错，更加简洁

    ListNode* rotateRight(ListNode* head, int k) {
        if(!head) return head;
        ListNode* p = head;
        int length = 1;
        while(p->next!=NULL){
            p = p->next;
            length++;
        }
        p->next = head;
        for(size_t i = 0;i<length-k%length;i++){
            p = p->next;
        }
        ListNode* result = p->next;
        p->next = NULL;
        return result;
    }

1、注意第一次while循环中的条件

2、求出链表长度之后，开始遍历的位置是尾部，以及遍历的长度是length-k。这点需要明确。

LeetCode82：消除重复元素2

ListNode* deleteDuplicates(ListNode* head) {
        ListNode* p = head;
        ListNode t(0);
        t.next = head;
        ListNode* before = &t;
        while(p!=NULL){
            int num = p->val;
            p = p->next;
            bool change = false;
            while(p!=NULL && p->val == num){
                change = true;
                p = p->next;
            }
            if(change){
                before->next = p;
            }else{
                before = before->next;
            }
        }
        return t.next;
    }

LeetCode82：消除重复元素

ListNode* deleteDuplicates(ListNode* head) {
        ListNode* p = head;
        ListNode t(0);
        t.next = head;
        while(p!=NULL){
            int num = p->val;
            ListNode* c = p;
            p = p->next;
            while(p!=NULL && p->val == num){
                p = p->next;
            }
            c->next = p;
            c = c->next;
        }
        return t.next;
    }

LeetCode86：分割链表

ListNode* partition(ListNode* head, int x) {
        ListNode t1(0);
        ListNode t2(0);
        ListNode* less = &t1;
        ListNode* bigger = &t2;
        ListNode* current = head;
        while(current!=NULL){
            if(current->val < x){
                less->next = current;
                less = less->next;
            }else{
                bigger->next = current;
                bigger = bigger->next;
            }
            current = current->next;
        }
        bigger->next = NULL;
        less->next = t2.next;
        return t1.next;
    }

1、注意退出while循环后将两个分别的链表拼起来的时候，大于val的链表尾部要设为空。

LeetCode206：颠倒链表

ListNode* reverseList(ListNode* head) {
        ListNode* before = NULL;
        while(head){
            ListNode* t = head->next;
            head->next = before;
            before = head;
            head = t;
        }
        return before;
    }

LeetCode206：颠倒链表2

ListNode t(0);
        t.next = head;
        ListNode* before_m;
        ListNode* _m;
        ListNode* _n;
        ListNode* later_n;
        ListNode* current = head;
        ListNode* before = &t;
        int num = 0;
        while(current){
            num++;
            if(num<m){
                before = current;
                current = current->next;
            }else if(num == m){
                before_m = before;
                _m = current;
                ListNode* t = current->next;
                current->next = NULL;
                before = current;
                current = t;
            }else if(num <= n){
                ListNode* t = current->next;
                current->next = before;
                before = current;
                current = t;
            }else{
                break;
            }
        }
        _n = before;
        later_n = current;
        
        before_m->next = _n;
        _m->next = later_n;
        return t.next;

完成整个大的目标需要两个条件

1、翻转m到n的链表

2、将before_m的next指向n，m的next指向later_n

然后遍历，在遍历过程中对几个过程进行处理，小于m时，等于m时，大于m小于等于n时

LeetCode109：将链表转化为二叉树

TreeNode* sortedListToBST(ListNode* head) {
        if(head == NULL) return NULL;
        if(head->next == NULL){
            TreeNode* result = new TreeNode(head->val);
            return result;
        }
        ListNode* before_l1 = NULL;
        ListNode* l1 = head;
        ListNode* l2 = head->next;
        while(l2!=NULL && l2->next!=NULL){
            before_l1 = l1;
            l1 = l1->next;
            l2 = l2->next->next;
        }
        TreeNode* node = new TreeNode(l1->val);
        if(before_l1 == NULL){
            node->left = sortedListToBST(NULL);
        }else{
            before_l1->next = NULL;
            node->left = sortedListToBST(head);
        }
        node->right = sortedListToBST(l1->next);
        return node;
    }

1、需要注意的是在将链表断成3个部分的时候前一部分有额外情况，那就是链表只有两个节点的时候，l1指向第一个节点，前一部分应该为NULL

LeetCode138：复制随机链表的指针

RandomListNode* copyRandomList(RandomListNode *head, map<int, RandomListNode*>& m) {
    if (head == NULL) return NULL;
    if (m.find(head->label) == m.end()) {
        RandomListNode* t = new RandomListNode(head->label);
        m.insert(pair<int, RandomListNode*>(head->label, t));
        t->next = copyRandomList(head->next, m);
        t->random = copyRandomList(head->random, m);
        return t;
    }
    else {
        return (*m.find(head->label)).second;
    }
}
RandomListNode *copyRandomList(RandomListNode *head) {
    if (head == NULL) return NULL;
    map<int,RandomListNode*> m;
    return copyRandomList(head,m);
}

LeetCode141 判断链表是否有环：

bool hasCycle(ListNode *head) {
        if(head == NULL) return false;
        ListNode t(0);
        t.next = head;
        ListNode* l1 = &t;
        ListNode* l2 = &t;
        while(l2!=NULL && l2->next!=NULL){
            l1 = l1->next;
            l2 = l2->next->next;
            if(l1 == l2) return true;
        }
        return false;
    }

1、判断l1等于l2的位置不能写错

LeetCode142 找出链表有环的位置

ListNode *detectCycle(ListNode *head) {
        if(!head) return  NULL;
        ListNode t(0);
        t.next = head;
        ListNode* l1 = head;
        ListNode* l2 = head;
        while(l2!=NULL && l2->next!=NULL){
            l1 = l1->next;
            l2 = l2->next->next;
            if(l1 == l2) break;
        }
        if(l2 == NULL || l2->next == NULL) return NULL;
        ListNode* l3 = head;
        while(l1 != l3){
            l1 = l1->next;
            l3 = l3->next;
        }
        return l1;
    }

1、本题链表可能没有环

LeetCode143 重新整理链表

void reorderList(ListNode* head) {
        if(head == NULL || head->next == NULL) return ;
        ListNode* l1 = head;
        ListNode* l2 = head->next;
        while(l2 != NULL && l2->next!=NULL){
            l1 = l1->next;
            l2 = l2->next->next;
        }
        ListNode* head2 = l1->next;
        l1->next = NULL;
        l1 = NULL;
        while(head2 != NULL){
            ListNode* t = head2->next;
            head2->next = l1;
            l1 = head2;
            head2 = t;
        }
        //l1是反向链表的头
        ListNode t(0);
        ListNode* result = &t;
        while(l1 != NULL || head!=NULL){
            if(head!=NULL){
                result->next = head;
                head = head->next;
                result = result->next;
            }
            if(l1 != NULL){
                result->next = l1;
                l1 = l1->next;
                result = result->next;
            }
        }
        head = t.next;
        return ;
    }

LeetCode 147 链表的插入排序

ListNode* insertionSortList(ListNode* head) {
    ListNode t(INT_MIN);
    t.next = NULL;
    ListNode* l1 = head;
    while (l1 != NULL) {
        //删除l1节点
        ListNode* next = l1->next;
        //从头寻找插入点
        ListNode* before_current = &t;
        ListNode* current = t.next;
        while (current != NULL && current->val<l1->val) {
            before_current = before_current->next;
            current = current->next;
        }
        l1->next = current;
        before_current->next = l1;


        l1 = next;
    }
    return t.next;
}

 LeetCode 148 链表的归并排序

ListNode* merge(ListNode* l1, ListNode* l2) {
    ListNode temp(-1);
    ListNode* result = &temp;
    while (l1 != NULL && l2 != NULL) {
        if (l1->val > l2->val) {
            result->next = l2;
            l2 = l2->next;
        }else {
            result->next = l1;
            l1 = l1->next;
        }
        result = result->next;
    }
    if (l1 == NULL)
        result->next = l2;
    else
        result->next = l1;
    return temp.next;
}
ListNode* sortList(ListNode* head) {
    if (head == NULL || head->next == NULL) return head;
    ListNode* l1 = head;
    ListNode* l2 = head->next;
    while (l2 != NULL && l2->next!= NULL) {
        l1 = l1->next;
        l2 = l2->next->next;
    }
    //l1是中间节点
    l2 = l1->next;
    l1->next = NULL;

    return merge(sortList(head), sortList(l2));
}

LeetCode160 找出两链表的交叉点

ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
        int lengthA = 0;
        ListNode* l = headA;
        while(l!=NULL){
            lengthA++;
            l = l->next;
        }
        int lengthB = 0;
        l = headB;
        while(l!=NULL){
            lengthB++;
            l = l->next;
        }
        ListNode* l1 = headA;
        ListNode* l2 = headB;
        if(lengthA<lengthB){
            int t = lengthB-lengthA;
            for(size_t i =0;i<t;i++)
                l2 = l2->next;
        }else{
            int t = lengthA-lengthB;
            for(size_t i =0;i<t;i++)
                l1 = l1->next;
        }
        while(l1 != l2){
            l1 = l1->next;
            l2 = l2->next;
        }
        return l1;
    }

LeetCode 203 删除节点中的元素

ListNode* removeElements(ListNode* head, int val) {
        if (head == NULL ) return NULL;
        ListNode* l = head;
        ListNode t(-1);
        t.next = head;
        ListNode* l_before = &t;
        while (l) {
            if (l->val == val)
                l_before->next = l->next;
            else
                l_before = l;
            l = l->next;
        }
        return t.next;
    }

LeetCode234 回文链表

bool isPalindrome(ListNode* head) {
        if(head == NULL || head->next == NULL) return true;
        ListNode* l1 = head;
        ListNode* l2 = head->next;
        while(l2!=NULL && l2->next!=NULL){
            l1 = l1->next;
            l2 = l2->next->next;
        }
        l2 = l1->next;
        l1->next = NULL;
        l1 = NULL;
        while(l2 != NULL){
            ListNode* t = l2->next;
            l2->next = l1;
            l1 = l2;
            l2 = t;
        }
        //l1是后一部分的头
        while(l1!=NULL && head!=NULL){
            if(l1->val != head->val) return false;
            l1 = l1->next;
            head = head->next;
        }
        return true;
    }

1、注意就情况的区分

LeetCode 237 删除节点

void deleteNode(ListNode* node) {
        if (!node || !node->next)
            return;
        node->val = node->next->val;
        node->next = node->next->next;
    }

LeetCode328 奇偶链表

ListNode* oddEvenList(ListNode* head) {
        if(head == NULL || head->next == NULL) return head;
        ListNode* head1 = head,*current1 = head;
        ListNode* head2 = head->next,*current2 = head->next;
        ListNode* current = head->next->next;
        while(current != NULL){
            current1->next = current;
            current1 = current1->next;
            current = current->next;
            if(current != NULL){
                current2->next = current;
                current2 = current2->next;
                current = current->next;
            }
        }
        current1->next = head2;
        current2->next = NULL;
        return head1;
    }

LeetCode445 反向相加两数

ListNode* reverse(ListNode* head){
        if(head == NULL || head->next == NULL) return head;
        ListNode* l1 = NULL;
        ListNode* l2 = head;
        while(l2){
            ListNode* t = l2->next;
            l2->next = l1;
            l1 = l2;
            l2 = t;
        }
        return l1;
    }
    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
        ListNode* new1 = reverse(l1);
        ListNode* new2 = reverse(l2);
        //加两个链表，前面有一个题和这里一模一样
        int add = 0;
        ListNode* current1 = new1;
        ListNode* current2 = new2;
        ListNode result(0);
        ListNode* current = &result;
        while(current1 || current2 || add==1){
            int num1 = 0;
            int num2 = 0;
            if(current1){
                num1 = current1->val;
                current1 = current1->next;
            } 
            if(current2) {
                num2 = current2->val;
                current2 = current2->next;
            }
            current->next = new ListNode((num1+num2+add)%10);
            add = (num1+num2+add)/10;
            current = current->next;
        }
        
        return reverse(result.next);
    }

LeetCode 430 铺展多级链表

Node* flatten(Node* head) {
        if(head == NULL || (head->child == NULL && head->next == NULL)) return head;
        if(head->child == NULL ) {
            head->next = flatten(head->next);
            head->next->prev = head;
            return head;
        }else if(head->next == NULL){
            head->next = flatten(head->child);
            head->next->prev = head;
            head->child = NULL;
            return head;
        }else{
            Node* n = flatten(head->next);
            Node* c = flatten(head->child);
            head->next = c;
            c->prev = head;
            Node* t = head;
            while(c->next!=NULL){
                c = c->next;
            }
            c->next = n;
            n->prev = c;
            head->child = NULL;
            return head;
        }
    }

这题和链表相关不大，会递归或是迭代的写法都可以。

三、复习题目

61

109

141

143

147

148

287