/******************************************************************************* 避免洪水泛滥 https://leetcode-cn.com/problems/avoid-flood-in-the-city/solution/avoid-flood-in-the-city-by-ikaruga/ rains : 1 0 2 0 2 1 *******************************************************************************/ class Solution { public: vector<int> avoidFlood(vector<int> &rains) { vector<int> ans(rains.size(), 1); unordered_map<int, int> water; //记录每个湖泊上一次下雨的日期 set<int> zero; for (int i = 0; i < rains.size(); i++) { int r = rains[i]; if (r == 0) { //将晴天的日期记录到zero中,遇到晴天时，先不用管抽哪个湖 zero.insert(i); continue; } if (water.count(r) != 0) { //当下雨时，湖泊已经水满时，可以查询上次下雨的日期 auto it = zero.lower_bound(water[r]); //通过上次下雨的日期，查找对应的晴天日期 if (it == zero.end()) { //如果没有找到，则无法使用那天抽水，发生洪水 return {}; } ans[*it] = r; // 如果在晴天*it找到了对应的下雨日期，则记录抽干的湖泊 zero.erase(it); //删除当前的晴天记录 } // water记录每个湖泊上一次下雨的日期 water[r] = i; //更新当前湖的下雨日期 ans[i] = -1; } return ans; } }; /******************************************************************************* 并查集 *******************************************************************************/ void join(int p, int q){ int rootP = find(p); int rootQ = find(q); if(rootP == rootQ){ return; } //将两棵树合并为一棵 parent[rootP] = rootQ; return; } int find(int p){ while(p != parent[p]){ //如果P的父亲指针指向不是自己，说明P不是根元素 //在find查询中嵌入一个路径压缩 此处不懂，参考这里https://blog.csdn.net/qq_19782019/article/details/78919990 parent[p] = parent[parent[p]]; //p元素不再选择原来的父亲节点，而是选择父亲节点的父亲节点作为自己的新的一个父亲节点 p = parent[p]; } //经过while循环后，p = parent[p], 一定是一个根节点，且不能够再进行压缩 return p; } /******************************************************************************* 1319. 连通网络的操作次数 https://leetcode-cn.com/problems/number-of-operations-to-make-network-connected/ 两种方法：并查集 和 深度优先探索 *******************************************************************************/ /*并查集*/ class Solution { public: int* parent; int find(int p){ while(p != parent[p]){ parent[p] = parent[parent[p]]; p = parent[p]; } return p; } void join(int x, int y){ int rootX = find(x); int rootY = find(y); if(rootX == rootY){ return; } parent[rootX] = rootY; return; } int makeConnected(int n, vector<vector<int>>& connections) { if(connections.size()<n-1) //不满足线缆数量 return -1; parent = new int[n]; //申请n个空间 for(int i =0;i<n;i++){ parent[i] = i; //初始化 } for(int i =0;i<connections.size();i++){ join(connections[i][0],connections[i][1]); //连接 } set<int> pars; for(int i =0;i<n;i++){ //计算连通分量数 pars.insert(find(i)); //每个节点的根节点插入pars，每个键都是唯一的，可以插入或删除但不能更改 } return pars.size()-1; } }; class Solution { private: vector<int> fa; public: int find(int x){ return x == fa[x] ? x : fa[x] = find(fa[x]); // 这种是未压缩路径 // return x == fa[x] ? x : x = find(fa[x]); } int makeConnected(int n, vector<vector<int>>& connections){ if(connections.size() < n-1){ return -1; } fa.resize(n); //初始化 iota(fa.begin(), fa.end(), 0); int part = n; //初始化共有n个连通分量数 for(auto&& c : connections){ //连接 int p = find(c[0]), q = find(c[1]); if(p != q){ --part; fa[p] = q; } } return part - 1; } }; /* 邻接矩阵深度优先探索算法 */ class Solution{ private: vector<vector<int>> edges; vector<bool> used; public: void dfs(int u){ used[u] = true; //将已访问的定点标记true for(int v : edges[u]){ //对于顶点u的其他相邻的节点，进行继续深度优先探索 if(!used[v]){ dfs(v); } } } int makeConnected(int n, vector<vector<int>> & connections){ if(connections.size() < n-1){ return -1; } edges.resize(n); for(auto&& c: connections){ //组合无向图 edges[c[0]].push_back(c[1]); edges[c[1]].push_back(c[0]); } used.resize(n); int part = 0; for(int i = 0; i < n; i++){ //遍历所有顶点，计算连通数量 if(!used[i]){ ++part; dfs(i); } } return part-1; } }; /*邻接矩阵的广度优先探索*/ class Solution { public: int makeConnected(int n, vector<vector<int>>& connections) { if (connections.size() < (n-1)) { return -1; } //将相连接的组合成一个无向图 unordered_map<int, unordered_set<int>> link_map; for (auto &conn : connections) { link_map[conn[0]].emplace(conn[1]); link_map[conn[1]].emplace(conn[0]); } vector<int> mark(n, 0); int num = 0; for (int i = 0; i < n; ++i) { if (mark[i]) continue; num++; queue<int> q; q.push(i); while (!q.empty()) { int index = q.front(); q.pop(); if (link_map.count(index) == 0) { continue; } for (auto mi : link_map[index]) { if (mark[mi]) continue; q.push(mi); mark[mi] = 1; } } } return num-1; } }; /******************************************************************************* 邻接矩阵：深度优先遍历，广度优先遍历 深度优先遍历：不断地沿着顶点的深度方向遍历；顶点的深度方向：它的邻接点方向。 用visited[i]表示顶点i的访问情况。 广度优先遍历：先访问当前顶点的所有邻接点；先访问顶点的邻接点先于后访问顶点的邻接点被访问 *******************************************************************************/ //访问标志数组 int visited[MAX] = {0}; void DFS(M G, int v){ visited[v] = 1; } /******************************************************************************* 朋友圈 https://leetcode-cn.com/problems/friend-circles/ *******************************************************************************/ class Solution { public: vector<int> fa; int find(int p){ while(p != fa[p]){ fa[p] = fa[fa[p]]; p = fa[p]; } return p; } void Union(int p, int q){ int rootP = find(p); int rootQ = find(q); if(rootP != rootQ){ fa[rootP] = rootQ; } return; } int findCircleNum(vector<vector<int>>& M) { int n = M.size(); fa.resize(n); //申请n个空间 for(int i = 0; i < n; i++){ //初始化 fa[i] = i; } //连接 for(int i=0;i<n;++i){ for(int j=0;j<n;++j){ if(M[i][j]==1 && i!=j) Union(fa[i], fa[j]); } } //计算有多少个连通分量数 int count = 0; for(int i = 0; i < n; i++){ if(fa[i] == i){ ++count; } } return count; //计算有多少个连通分量数 // set<int> pars; // for(int i =0;i<n;i++){ // pars.insert(find(i)); //每个节点的根节点插入pars，每个键都是唯一的，可以插入或删除但不能更改 // } // return pars.size(); } }; /*深度优先探索*/ //把矩阵 M 看成是图的邻接矩阵，则问题转化为求图的连通块数 class Solution { private: vector<bool> vis; public: int findCircleNum(vector<vector<int>>& M){ int n = M.size(); vis.resize(n); int ans = 0; for(int i = 0; i < n; i++){ if(vis[i] == false){ dfs(M, i); ans++; } } return ans; } void dfs(vector<vector<int>> M, int i){ int n = M.size(); for(int j = 0; j < n; j++){ if(M[i][j] == 1 && vis[j] == false){ vis[j] = true; dfs(M, j); } } } }; /*广度优先探索*/ class Solution { public: int findCircleNum(vector<vector<int>>& M){ int n = M.size(); vector<int> visit(n, 0); int count = 0, tmp; queue<int> que; for(int i = 0; i < n; i++){ if(!visit[i]){ count++; que.push(i); while(!que.empty()){ tmp = que.front(); que.pop(); visit[tmp] = 1; for(int j = 0; j < n; j++){ if(M[tmp][j] == 1 && !visit[j]){ que.push(j); } } } } } return count; } }; /******************************************************************************* 820. 单词的压缩编码 https://leetcode-cn.com/problems/short-encoding-of-words/ *******************************************************************************/ class Solution { public: int minimumLengthEncoding(vector<string>& words) { //unordered_set<string> good(words.begin(), words.end()); unordered_set<string> origin; for(auto& word : words){ origin.insert(word); } for (const string& word: words) { for (int k = 1; k < word.size(); ++k) { origin.erase(word.substr(k)); } } int ans = 0; for (const string& word: origin) { ans += word.size() + 1; } return ans; } }; /******************************************************************************* LCP 08. 剧情触发时间 https://leetcode-cn.com/problems/ju-qing-hong-fa-shi-jian/ *******************************************************************************/ #define col 3 class Solution { public: vector<int> getTriggerTime(vector<vector<int>>& increase, vector<vector<int>>& requirements) { vector<int> res; int row = increase.size(); cout << row << endl; vector<vector<int>> sum(row+1, vector<int>(col,0)); for(int i = 0; i < row; i++){ for(int j = 0; j < col; j++){ sum[i+1][j] = sum[i][j] + increase[i][j]; } } for(auto v : requirements){ int l = 0, r = row; while(l < r){ int m = (l + r) / 2; if(sum[m][0] >= v[0] && sum[m][1] >= v[1] && sum[m][2] >= v[2]){ r = m; } else { l = m + 1; } } if(sum[l][0] >= v[0] && sum[l][1] >= v[1] && sum[l][2] >= v[2]){ res.push_back(l); } else { res.push_back(-1); } } return res; } }; class Solution{ public: vector<int> getTriggerTime(vector<vector<int>>& increase, vectoro<vector<int>>& requirements){ vector<int> C(increase.size()+1, 0); vector<int> R(increase.size()+1, 0); vector<int> H(increase.size()+1, 0); for(int i = 0; i < inrease.size(); ++i){ C[i+1] = C[i] + increase[i][0]; R[i+1] = R[i] + increase[i][1]; H[i+1] = H[i] + increase[i][2]; } vector<int> res; int maxLen = C.size(); for(int i = 0; i < requirements.size(); i++){ int lc = lower_bound(C.begin(), C.end(), requirements[i][0]) - C.begin(); int lr = lower_bound(R.begin(), R.end(), requirements[i][1]) - R.begin(); int lh = lower_bound(H.begin(), H.end(), requirements[i][2]) - H.begin(); if(lc == maxLen || lr == maxLen || lh == maxLen){ res.push_back(-1); } else { res.push_back(max(max(lc,lr),lh)); } } return res; } }; /******************************************************************************* 76. 最小覆盖子串 https://leetcode-cn.com/problems/minimum-window-substring/ *******************************************************************************/ class Solution { public: string minWindow(string s, string t) { unordered_map<char, int> need; unordered_map<char, int> windows; int len = s.size() + 1; int start = 0; //先计算need for(auto w : t){ ++need[w]; } int left = 0, right = 0, valid = 0; //初始化 while(right < s.size()){ char chr = s[right]; ++right; if(need.count(chr)){ ++windows[chr]; if(windows[chr] == need[chr]){ valid++; } } while(valid == need.size()){ if(right - left < len){ start = left; len = right - left; } char chl = s[left]; ++left; if(need.count(chl)){ if(windows[chl] == need[chl]){ --valid; } --windows[chl]; } } } return len == s.size()+1 ? "" : s.substr(start, len); } }; /******************************************************************************* 438. 找到字符串中所有字母异位词 https://leetcode-cn.com/problems/find-all-anagrams-in-a-string/ *******************************************************************************/ class Solution { public: vector<int> findAnagrams(string s, string p) { unordered_map<char, int> window; unordered_map<char, int> need; for(auto c : p){ need[c]++; } int left = 0, right = 0, valid = 0; vector<int> res; while(right < s.size()){ char c = s[right]; right++; if(need.count(c)){ window[c]++; if(window[c] == need[c]){ valid++; } } //判断左侧端口是否要收缩 while(right - left >= p.size()){ if(valid == need.size()){ res.push_back(left); } char d = s[left]; left++; if(need.count(d)){ if(window[d] == need[d]){ valid--; } window[d]--; } } } return res; } }; /******************************************************************************* 567. 字符串的排列 https://leetcode-cn.com/problems/permutation-in-string/ *******************************************************************************/ class Solution { public: bool checkInclusion(string s1, string s2) { unordered_map<char, int> need, window; for(auto c : s1){ need[c]++; } int left = 0, right = 0; int valid = 0; while(right < s2.size()){ char c = s2[right]; right++; if(need.count(c)){ window[c]++; if(window[c] == need[c]){ valid++; } } while(right - left >= s1.size()){ if(valid == need.size()){ return true; } char d = s2[left]; left++; if(need.count(d)){ if(window[d] == need[d]){ valid--; } window[d]--; } } } return false; } }; /******************************************************************************* 3. 无重复字符的最长子串 https://leetcode-cn.com/problems/longest-substring-without-repeating-characters/ *******************************************************************************/ class Solution { public: int lengthOfLongestSubstring(string s) { unordered_map<char, int> window; int left = 0, right = 0; int res = 0; while(right < s.size()){ char c = s[right]; right++; //进行窗口数据更新 window[c]++; //判断窗口是否要收缩 while(window[c] > 1){ char d = s[left]; left++; window[d]--; } res = max(res, right-left); } return res; } }; /******************************************************************************* II. 左旋转字符串 https://leetcode-cn.com/problems/zuo-xuan-zhuan-zi-fu-chuan-lcof/ *******************************************************************************/ class Solution { public: string reverseLeftWords(string s, int n) { reverse(s.begin(), s.begin()+n); reverse(s.begin()+n, s.end()); reverse(s.begin(), s.end()); return s; } }; /******************************************************************************* N皇后 https://leetcode-cn.com/problems/n-queens/ *******************************************************************************/ class Solution { public: vector<vector<string>> res; vector<vector<string>> solveNQueens(int n) { vector<string> board(n, string(n, '.')); backtrack(board, 0); return res; } void backtrack(vector<string>& board, int row){ //触发条件 if(row == board.size()){ res.push_back(board); return; } int n = board[row].size(); for(int col = 0; col < n; col++){ if(!isValid(board ,row, col)){ continue; } //做选择 board[row][col] = 'Q'; // 进行下一步策略 backtrack(board, row+1); //撤回选择 board[row][col] = '.'; } } bool isValid(vector<string>& board, int row, int col){ int n = board.size(); //检查列是否有皇后冲突 for(int i = 0; i < n; i++){ if(board[i][col] == 'Q'){ return false; } } //检查右上方是否有皇后互相冲突 for(int i = row-1, j = col + 1; i>=0 && j < n; i--, j++){ if(board[i][j] == 'Q'){ return false; } } //检查左上方是否有皇后互相冲突 for(int i = row-1, j = col - 1; i>=0 && j >= 0; i--, j--){ if(board[i][j] == 'Q'){ return false; } } return true; } }; /******************************************************************************* 980. 不同路径 III https://leetcode-cn.com/problems/unique-paths-iii/ *******************************************************************************/ class Solution { public: int uniquePathsIII(vector<vector<int>>& grid) { int startX = 0, startY = 0, stepNum = 1; //遍历获取起始位置和统计总步数 for(int i = 0; i < grid.size(); i++){ for(int j = 0; j < grid[0].size(); j++){ if(grid[i][j] == 1){ startX = j; startY = i; continue; } if(grid[i][j] == 0){ stepNum++; } } } return dfs(startX, startY, stepNum, grid); } int dfs(int x, int y, int stepSur, vector<vector<int>>& grid){ if(x < 0 || x >= grid[0].size() || y < 0 || y >= grid.size() || grid[y][x] == -1){ return 0; } if (grid[y][x] == 2) return stepSur == 0 ? 1 : 0; grid[y][x] = -1; int res = 0; res += dfs(x-1, y, stepSur -1 ,grid); res += dfs(x + 1, y, stepSur - 1, grid); res += dfs(x, y - 1, stepSur - 1, grid); res += dfs(x, y + 1, stepSur - 1, grid); grid[y][x] = 0; return res; } }; /******************************************************************************* 833. 字符串中的查找与替换 https://leetcode-cn.com/problems/find-and-replace-in-string/ *******************************************************************************/ class Solution { public: string findReplaceString(string S, vector<int>& indexes, vector<string>& sources, vector<string>& targets) { int n = S.size(); vector<string> vec(n, ""); string res = ""; for (int i = 0; i < S.size(); ++i) vec[i] += S[i]; for (int i = 0; i < indexes.size(); ++i) { int size = sources[i].size(); if (S.substr(indexes[i], size) == sources[i]) { vec[indexes[i]] = targets[i]; for (int j = indexes[i] + 1; j < indexes[i] + size; ++j) vec[j] = ""; } } for (auto str : vec) { res += str; } return res; } }; /******************************************************************************* 84. 柱状图中最大的矩形 https://leetcode-cn.com/problems/largest-rectangle-in-histogram/ *******************************************************************************/ class Solution { public: int largestRectangleArea(vector<int>& heights) { int ans = 0; vector<int> st; heights.insert(heights.begin(), 0); heights.push_back(0); for (int i = 0; i < heights.size(); i++) { while (!st.empty() && heights[st.back()] > heights[i]) { int cur = st.back(); st.pop_back(); int left = st.back() + 1; int right = i - 1; ans = max(ans, (right - left + 1) * heights[cur]); } st.push_back(i); } return ans; } }; class Solution { public: int largestRectangleArea(vector<int>& heights){ int res = 0; stack<int> tack; heights.insert(heights.begin(), 0); heights.push_back(0); for(int i = 0; i < heights.size(); i++){ while(!tack.empty() && heights[tack.top()] > heights[i]){ int cur = tack.top(); tack.pop(); int left = tack.top() + 1; int right = i - 1; //heights[cur] 是 离heights[i] 最远的那个且大于heights[i]，right-left 是递增的宽度 res = max(res, (right - left + 1) * heights[cur]); } tack.push(i); } return res; } }; /******************************************************************************* 735. 行星碰撞 https://leetcode-cn.com/problems/asteroid-collision/ *******************************************************************************/ class Solution { public: vector<int> asteroidCollision(vector<int>& asteroids) { vector<int> res; for(auto v : asteroids){ bool flag = true; while(!res.empty() && v < 0 && res.back() > 0){ int sum = res.back() + v; if(sum <= 0){ res.pop_back(); } if(sum >= 0) { flag = false; break; } } if(flag){ res.push_back(v); } } return res; } }; /******************************************************************************* 2. 两数相加 https://leetcode-cn.com/problems/add-two-numbers/ *******************************************************************************/ class Solution { public: ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) { ListNode* head = nullptr; ListNode* tail = nullptr; int carry = 0; if(l1 == nullptr){ return l2; } if(l2 == nullptr){ return l1; } while(l1 || l2){ int n1 = l1 ? l1->val : 0; int n2 = l2 ? l2->val : 0; int sum = n1 + n2 + carry; if(!head){ head = tail = new ListNode(sum % 10); } else { tail->next = new ListNode(sum % 10); tail = tail->next; } carry = sum / 10; if(l1){ l1 = l1->next; } if(l2){ l2 = l2->next; } } if(carry > 0){ tail->next = new ListNode(carry); } return head; } }; /******************************************************************************* 15. 三数之和---左右指针 https://leetcode-cn.com/problems/3sum/ *******************************************************************************/ class Solution { public: vector<vector<int>> threeSum(vector<int>& nums) { vector<vector<int>> res; sort(nums.begin(), nums.end()); int target; for(int i = 0; i < nums.size(); i++){ if(i > 0 && nums[i] == nums[i-1]){ continue; } if((target = nums[i]) > 0){ break; } int l = i + 1; int r = nums.size()-1; while(l < r){ if(nums[l] + target + nums[r] < 0){ ++l; } else if(nums[l] + target + nums[r] > 0){ --r; } else { res.push_back({nums[l], target, nums[r]}); ++l; --r; while(l < r && nums[l] == nums[l-1]){ l++; } while(l < r && nums[r] == nums[r+1]){ r--; } } } } return res; } }; /******************************************************************************* 面试题 08.12. 八皇后---回溯法 https://leetcode-cn.com/problems/eight-queens-lcci/ *******************************************************************************/ class Solution1 { public: vector<vector<string>> res; vector<vector<string>> solveNQueens(int n) { vector<string> board(n, string(n, '.')); backtrack(board ,0); return res; } void backtrack(vector<string>& board, int row){ if(row == board.size()){ res.push_back(board); return; } int n = board[row].size(); for(int col = 0; col < n; col++){ if(!isValid(board, row, col)){ continue; } board[row][col] = 'Q'; backtrack(board, row+1); board[row][col] = '.'; } } bool isValid(vector<string>& board, int row, int col){ int n = board.size(); //检查列是否有冲突 for(int i = 0; i < n; i++){ if(board[i][col] == 'Q'){ return false; } } //检查右上方对角线 for(int i = row-1, j= col+1; i >= 0 && j < n; i--, j++){ if(board[i][j] == 'Q'){ return false; } } //检查左上方对角线 for(int i = row-1, j = col-1; i >= 0 && j >= 0; i--, j--){ if(board[i][j] == 'Q'){ return false; } } return true; } }; /******************************************************************************* 980. 不同路径 III---回溯法 https://leetcode-cn.com/problems/unique-paths-iii/ *******************************************************************************/ class Solution3{ public: int uniquePathsIII(vector<vector<int>>& grid){ int res = 0; int startX = 0, startY = 0; int step = 0; for(int i = 0; i < grid.size(); i++){ for(int j = 0; j < grid[0].size(); j++){ if(grid[i][j] == 1){ startX = i; startY = j; continue; } if(grid[i][j] == 0){ step++; } } } //step + 1 针对step != 0 时，则直接返回res = 0 trackBack(grid, startX, startY, step+1, res); return res; } void trackBack(vector<vector<int>>& grid, int startX, int startY, int step, int &res){ if(startX < 0 || startX >= grid.size() || startY < 0 || startY >= grid[startX].size() || grid[startX][startY] == -1){ return; } if(grid[startX][startY] == 2){ if(step == 0){ res++; } return; } grid[startX][startY] = -1; trackBack(grid, startX-1, startY, step-1, res); trackBack(grid, startX+1, startY, step-1, res); trackBack(grid, startX, startY-1, step-1, res); trackBack(grid, startX, startY+1, step-1, res); grid[startX][startY] = 0; } };