三數之和
https://leetcode.cn/problems/3sum/description/
給你一個整數數組 nums ，判斷是否存在三元組 [nums[i], nums[j], nums[k]] 滿足 i != j、i != k 且 j != k ，同時還滿足 nums[i] + nums[j] + nums[k] == 0 。請你返回所有和為 0 且不重復的三元組。
注意：答案中不可以包含重復的三元組。
其實仔細想想，三個指針里面i是單向移動的？？ 還可以簡化
class Solution {
public:vector<vector<int>> threeSum(vector<int>& nums) {vector<vector<int>> ans;std::sort(nums.begin(), nums.end());if (nums.size() == 0 || nums[0] > 0) {return ans;}int nums_len = nums.size();for (int left = 0; left < nums_len; left++) {int target;int last_num = -nums[left];if (left > 0 && nums[left] == nums[left - 1]) {continue;}int right = nums_len - 1;target = -nums[left];for (int i = left + 1; i < nums_len; i++) {if (i > left + 1 && nums[i] == nums[i - 1]) {continue;}while (i < right && nums[i] + nums[right] > target) {--right;}if (right == i) {break;}if (target == nums[i] + nums[right]) {ans.push_back({nums[left], nums[i], nums[right]});}}}return ans;}
};
vector<vector<int>> threeSum(vector<int>& nums) {int nums_len = nums.size();sort(nums.begin(), nums.end(), [](int a, int b) { return a < b; });vector<vector<int>> ans;for (int left = 0; left < nums_len; left++) {if (left > 0 && nums[left] == nums[left - 1]) {// 遇到重復的進行跳過continue;}int right = nums_len - 1;for (int mid = left + 1; mid < nums_len; mid++) {if (mid > left + 1 && nums[mid] == nums[mid - 1]) {// 遇到重復的進行跳過continue;}while (mid < right &&nums[mid] + nums[left] + nums[right] > 0) {right--;}if (right == mid) {break;}if (nums[mid] + nums[left] + nums[right] == 0) {ans.push_back({nums[left], nums[mid], nums[right]});}}}return ans;}
)
