LeetCode（M）threesum

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]
]

class Solution {public:    vector<vector<int>> threeSum(vector<int>& nums) {        if(nums.size()<3) return{};        vector<vector<int>> v;        sort(nums.begin(),nums.end());        for(int i=0;i<nums.size();++i){            int target=-nums[i];            if(target<0) break  ;                int j=i+1,k=nums.size()-1;                while(j<k){                    if(nums[j]+nums[k]==target){                       v.push_back({nums[i],nums[j],nums[k]});                    //   while(j+1<k&&nums[j]==nums[j+1]) j++;while(j<k-1&&nums[k]==nums[k-1]) k--;//point1             int v1=nums[j],v2=nums[k]; while(j<k&&nums[j]==v1) j++;while(j<k&&nums[k]==v2) k--;//point2                    }                    if(nums[j]+nums[k]<target) ++j;                    if(nums[j]+nums[k]>target)  --k;            }            while(i+1<nums.size()&&nums[i]==nums[i+1]) i++;//pointA                    }                    return v;    }};

在LintCode上写过这题，http://blog.csdn.net/jang1996/article/details/53780446
这次，在leetCode上超时了，加入point2、pointA两句才过。最开始我将point2写成point1，犯错了。
point1的意思是前一个数与后一个数相等，那么++或–；point2的意思是，在后面的数里，凡是和nums[j]或nums[k]相等的，都跳过，以免在v里重复。