3sum

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:

•Elements in a triplet (a,b,c) must be in non-descending order. (ie, a ≤ b ≤ c)
•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)

class Solution {
public:
    vector<vector<int>> threeSum(vector<int>& nums) {
        sort(nums.begin(),nums.end());
        vector<vector<int>> ret;
        for(int i = 0;i < nums.size();i++)
        {
            int start =  i+1;
            int end = nums.size()-1;
            if(i>0&&nums[i]==nums[i-1])//关键点1：去除已经反复的点
                continue;
            while(start<end)//转化为2sum问题
            {

                while(start>i+1&&nums[start]==nums[start-1])//关键点2：排除反复的部分
                {
                    start++;
                }
                while(end<nums.size()-1&&nums[end]==nums[end+1])
                {
                    end--;
                }
                int sum = nums[start]+nums[i]+nums[end];
                if(start<end)
                {
                    if(sum == 0)
                    {
                        vector<int> temp;
                        temp.push_back(nums[i]);
                        temp.push_back(nums[start]);
                        temp.push_back(nums[end]);
                        ret.push_back(temp);
                        start++;
                    }else if(sum>0)
                    {

                        end--;
                    }else{

                        start++;
                    }
                } 
            }

        }
        return ret;
    }
};