LeetCode--3Sum

类似于2sum，先排序，然后从左开始遍历，计算a[i]后面的等于-a[i]的两个元素，

注意去除重复元素

class Solution {
public:
    vector<vector<int> > threeSum(vector<int> &num) {
        vector<vector<int> > res;
        sort(num.begin(),num.end());
        int n = num.size();
        for(int i = 0 ; i < n-2 ; ++i){
            if(i > 0 && (num[i] == num[i-1])){
                continue;
            }
            int target = -num[i];
            int left = i+1;
            int right = n-1;
            while(left < right){
                if(target == num[left] + num[right]){
                    vector<int> triplet;
                    triplet.push_back(num[i]);
                    triplet.push_back(num[left]);
                    triplet.push_back(num[right]);
                    res.push_back(triplet);
                    left++;
                    while((left < right) && (num[left] == num[left-1])){
                        left++;
                    }
                    right--;
                    while((right > left) && (num[right] == num[right+1])){
                        right--;
                    }
                }
                else if(target > num[left] + num[right]){
                    left++;
                    while((left < right) && (num[left] == num[left-1])){
                        left++;
                    }
                }
                else{
                    right--;
                    while((right > left) && (num[right] == num[right+1])){
                        right--;
                    }
                }
            }
        }
        return res;
    }
};