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@ -61,6 +61,8 @@ public:
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vector<vector<int>> threeSum(vector<int>& nums) {
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vector<vector<int>> threeSum(vector<int>& nums) {
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vector<vector<int>> result;
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vector<vector<int>> result;
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sort(nums.begin(), nums.end());
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sort(nums.begin(), nums.end());
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// 找出a + b + c = 0
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// a = nums[i], b = nums[left], c = nums[right]
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for (int i = 0; i < nums.size(); i++) {
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for (int i = 0; i < nums.size(); i++) {
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// 排序之后如果第一个元素已经大于零,那么无论如何组合都不可能凑成三元组,直接返回结果就可以了
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// 排序之后如果第一个元素已经大于零,那么无论如何组合都不可能凑成三元组,直接返回结果就可以了
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if (nums[i] > 0) {
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if (nums[i] > 0) {
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@ -1,4 +1,3 @@
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## 题目地址
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## 题目地址
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https://leetcode-cn.com/problems/valid-anagram/
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https://leetcode-cn.com/problems/valid-anagram/
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19
problems/0344.反转字符串.md
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19
problems/0344.反转字符串.md
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@ -0,0 +1,19 @@
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## 题目地址
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https://leetcode-cn.com/problems/reverse-string/
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## 思路
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遍历数组的前一半,同时和后一半做交换就可以了
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## 代码
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```
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class Solution {
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public:
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void reverseString(vector<char>& s) {
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for (int i = 0, j = s.size() - 1; i < s.size()/2; i++, j--) {
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swap(s[i],s[j]);
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}
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}
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};
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```
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