3Sum

Given an array nums of n integers, are there elements a, b, c in nums 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.

Example:

Given array nums = [-1, 0, 1, 2, -1, -4],

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

class Solution {
public:
    vector<vector<int>> twoSum(vector<int>& nums,int start,int target){
        vector<vector<int>> ans;
        int i=start,j=nums.size()-1;
        while(i<j)
        {
            int sum=nums[i]+nums[j];
            int left =nums[i],right = nums[j];
            if(sum == target)
            {
                ans.push_back({left,right});
                while(i<j && nums[i] == left) i++;//跳过重复的元素
                while(i<j &&nums[j] == right) j--;
            }
            else if(sum<target)
            {
                while(i<j && nums[i] == left) i++;
            }
            else
            {
                while(i<j &&nums[j] == right) j--;
            }
        }
        return ans;
    }
    vector<vector<int>> threeSum(vector<int>& nums) {
        sort(nums.begin(),nums.end());
        int n = nums.size();
        vector<vector<int>> ans;
        for(int i=0;i<n;i++)
        {
            vector<vector<int>> tuples = twoSum(nums,i+1,-nums[i]);//找出符合的二元组
            for(vector<int>& tuple :tuples)
            {
                tuple.push_back(nums[i]);
                ans.push_back(tuple);
            }
            // 跳过第一个数字重复的情况，否则会出现重复结果
            while (i < n - 1 && nums[i] == nums[i + 1]) i++;
        }
        return ans;
    }
};

注意：

1.双指针法，三个数的的合就先枚举每个数，然后再用双指针找另外两个数。

2.注意去重，连续相同的元素要及时跳过，在两个数的和时就要注意跳过。