lintcode:3 Sum

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.

Example
For example, given array S = {-1 0 1 2 -1 -4}, A solution set is:

(-1, 0, 1)
(-1, -1, 2)
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.

class Solution {
public:    
    /**
     * @param numbers : Give an array numbers of n integer
     * @return : Find all unique triplets in the array which gives the sum of zero.
     */
    vector<vector<int> > threeSum(vector<int> &nums) {
        // write your code here
        sort(nums.begin(),nums.end());

        vector<vector<int> > res;

        for(int i=0;i<nums.size();i++){

           if(i>=1&&nums[i]==nums[i-1]){
                continue;
            }

            int pa=i+1;
            int pb=nums.size()-1;

            bool first=true;

            int sum=0;

            while(pa<pb){


                sum=nums[i]+nums[pa]+nums[pb];

                if(sum==0){

                    if(!first&&nums[pa]==nums[pa-1]){
                       pa++;
                       continue;
                    }
                    if(!first&&nums[pb]==nums[pb+1]){
                       pb--;
                       continue;
                    }

                    vector<int> tmp;
                    tmp.push_back(nums[i]);
                    tmp.push_back(nums[pa]);
                    tmp.push_back(nums[pb]);

                    res.push_back(tmp);

                    pa++;
                    pb--;

                    if(first){
                      first=false;
                    }


                }else if(sum<0){
                    pa++;
                }else{
                    pb--;
                }



            }

        }
        return res;
    }
};

跟3sum closest类似
注意排除重复triplets的方式