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上寫過這題，http://blog.csdn.net/jang1996/article/details/53780446
這次，在leetCode上超時了，加入point2、pointA兩句才過。最開始我將point2寫成point1，犯錯了。
point1的意思是前一個數與後一個數相等，那麼++或–；point2的意思是，在後面的數裏，凡是和nums[j]或nums[k]相等的，都跳過，以免在v裏重複。