Edward_LF的博客

class Solution {public: int divide(int x, int y) { using ll = long long; vector<ll> exp; bool is_minus = false; if(x < 0 && y >0 || x > 0 && y < 0) is_minus = true; ll a = ..

解题思路：双指针；class Solution {public: int removeElement(vector<int>& nums, int val) { int k = 0; for(int i = 0; i < nums.size(); i++) { if(nums[i] != val) { nums[k++] = nums[.

解题思路：双指针做法class Solution {public: int removeDuplicates(vector<int>& nums) { int k = 0; for(int i = 1; i < nums.size(); i++) { if(nums[i] != nums[i-1]) { nums[++k] = num.

class Solution {public: ListNode* reverseKGroup(ListNode* head, int k) { auto dummy = new ListNode(-1); dummy->next = head; for(auto p = dummy;;) { //每次循环，判断后面是否够k个 auto q = p; ..

22. 括号生成括号问题思路：（通用）任意前缀中，左括号数量 >= 右括号数量；左右括号数量相等；使用dfs：左括号数量 <= n即可；右括号数量 <= n && 左括号数量 > 右括号数量class Solution {private: vector<string> res;public: vector<string> generateParenthesis(int n) {

解题思路：使用dummy节点。class Solution {public: ListNode* mergeTwoLists(ListNode* list1, ListNode* list2) { auto dummy = new ListNode(-1), tail = dummy; while(list1 && list2) { if(list1->val < list2->v.

解题思路：使用栈；使用ascll码进行判断，匹配括号的ascll码最多差2；class Solution {public: bool isValid(string s) { stack<char> sk; for(auto& c : s) { if(c == '(' || c == '{' || c == '[') sk.push(c); else .

解题思路：定义一个dummy指针；（要考虑头结点，最好是有dummy节点）找到要删除的节点的前一个节点—也就是导师k+1个节点。class Solution {public: ListNode* removeNthFromEnd(ListNode* head, int n) { auto dummy = new ListNode(-1); dummy->next = head; int k = 0; for(aut.

双指针；class Solution {public: vector<vector<int>> threeSum(vector<int>& nums) { vector<vector<int>> res; sort(nums.begin(), nums.end()); for(int i = 0; i < nums.size(); i++) { .

双指针；class Solution {public: int threeSumClosest(vector<int>& nums, int target) { sort(nums.begin(), nums.end()); pair<int,int> res(INT_MAX, INT_MAX); for(int i = 0; i < nums.size(); i++) { .

双指针做法；class Solution {public: vector<vector<int>> fourSum(vector<int>& nums, int target) { sort(nums.begin(), nums.end()); vector<vector<int>> res; for (int i = 0; i < nums.size(); i++) .

LeetCode17. 电话号码的字母组合dfs问题class Solution {private: vector<string> ans; const string str[10] = { "","", "abc","def", "ghi","jkl", "mno","pqrs", "tuv","wxyz" };public: vector<string