end()) { std::cout << "Value found in vector: " << *it << "\n"; } else { std::cout << "Value not found in vector.\n"; } return 0; } 注意的事 queue 的 pop C++ 中,空队列上调用 pop() 函数时会引发 undefined behavior (segmentation fault). 需要访问删除元素要使用 front(...
0251 Flatten 2D Vector 43.90% Medium 0252 Meeting Rooms 52.30% Easy 0253 Meeting Rooms II 43.10% Medium 0254 Factor Combinations 44.50% Medium 0255 Verify Preorder Sequence in Binary Search Tree 43.70% Medium 0256 Paint House 49.10% Easy 0257 Binary Tree Paths Go 46.30% Easy 0258 Ad...
templateTUVW//1. 最终结果保存,所有解的集合vector<vector<U>>res;//2. 访问标记(可选)//3. 字典(可选),这个字典是in元素变换到out元素集合的映射dict//4. 遍历优化(可选),二维遍历时的方向优化//5. 递归函数实现//其中in是输入,out是最终输出结果集合的每一个结果(在终止条件前是结果的部分)//start...
3. int evalRPN(vector<string> &tokens) { 4. <string> s; 5. for (auto token : tokens) 6. { 7. if (!is_operator(token)) 8. { 9. .push(token); 10. } 11. else 12. { 13. int y = stoi(s.top()); 14. .pop(); 15. int x = stoi(s.top()); 16. .pop(); 17. ...
nums = [1,3,-1,-3,5,3,6,7] k = 3 -> [3,3,5,5,6,7] Sliding window deque, ensure monotonic decr, leftmost largest Time: O(n) Space: O(n) */ class Solution { public: vector<int> maxSlidingWindow(vector<int>& nums, int k) { deque<int> dq; vector<int> result; int ...
1. vector; first,lastInput iterators to the initial and final positionsina range. The range used is [first,last),whichincludes all the elements between first andlast, including the element pointed by first but not the element pointed bylast. ...
Therefore it is not a palindrome. Example 3: Input: 10 Output: false Explanation: Reads 01 from right to left. Therefore it is not a palindrome. Follow up: Coud you solve it without converting the integer to a string? 10 package leetcode import strconv func isPalindrome(x int) bool {...
template<typename _Callable, typename... _Args, typename = _Require<__not_same<_Callable>>> explicit thread(_Callable&& __f, _Args&&... __args) { static_assert( __is_invocable<typename decay<_Callable>::type, typename decay<_Args>::type...>::value, "std::thread arguments must be...
If both characters are not the same. We can delete character from either string. So the min no of moves has to be minimum between dp[i][j - 1] and dp[i - 1][j]. Code: intminDistance(stringword1,stringword2){intm = word1.length();intn = word2.length();vector...
vector<TreeNode*>& out, int& res==>unordered_map<int, int>& m 然后看子路径之和有没有等于给定值的。 key的意义是 cur - sum,当 cur - sum == 0时,对应的结果是1个, 即开始就设置m[0]=1; 记忆搜索,主要思想:记录深度搜索时——从根节点往下的累积和(记为当前和cur),并每次将cur出现次数+...