【LeeCode】 15. 3Sum 解题小结

题目：

Given an array S of n integers, are there elements a, b, c in S such that a + b + c = 0? Find all unique triplets in the array which gives the sum of zero.

Note: The solution set must not contain duplicate triplets.

For example, given array S = [-1, 0, 1, 2, -1, -4],

A solution set is:
[
  [-1, 0, 1],
  [-1, -1, 2]
]

这题比较麻烦的点在于如何忽略重复元素的影响。首先将元素排序，然后从最左边开始，如果有三个元素相加为0，那么就把这三个数push进结果res里，然后排除重复元素。如果a+b+c的和大于0，那么从右边缩小范围，如果小于0，从左边缩小范围。

class Solution {
public:
    vector<vector<int>> threeSum(vector<int>& nums) {
        vector<vector<int>> res; 
        
        if (nums.size()<3) return res;
        sort(nums.begin(), nums.end());
        for (int i = 0; i < nums.size() - 2; i++){
            if (i > 0 && nums[i] == nums[i-1])continue;
            int a = nums[i];
            int left = i+1;
            int right = nums.size()-1;
            while(left<right){
                int b = nums[left];
                int c = nums[right];
                if ((a+b+c) == 0) {
                    res.push_back(vector<int>{a,b,c});
                    while((left < nums.size()) && (nums[left] == b))left++;
                    while((right >= 0) && (nums[right] == c))right--;
                }
                else if ((a+b+c)>0) right--;
                else left++;
            }
        }
        return res;
    }
};