3Sum

时间：2014-10-19 19:50:58 收藏：0 阅读：207

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)

分析：该题的主要思路是：先对num排序，然后最外层循环从第一个元素到倒数第三个元素，循环里用2sum两个指针的解法。此题的难点在于如何remove duplicates。如果在最后对结果除重，会超时，所以要在产生的过程中防止产生重复的triplet。

class Solution {
public:
    vector<vector<int> > threeSum(vector<int> &num) {
        vector<vector<int>> res;
        int n = num.size();
        if(n < 3) return res;
        
        sort(num.begin(), num.end());
        for(int i = 0; i < n-2; i++){
　　　　　　　//remove duplicates
            if(i == 0 || (i > 0 && num[i-1] != num[i])){
                int left = i+1, right = n-1;
                while(left < right){
                    if(num[i] + num[left] + num[right] < 0)
                        left++;
                    else if(num[i] + num[left] + num[right] > 0)
                        right--;
                    else{
                        res.push_back({num[i], num[left], num[right]});
                        left++;
                        right--;
　　　　　　　　　　　　　　//remove duplicates
                        while(left < right && num[left] == num[left-1])
                            left++;
                        while(left < right && num[right] == num[right+1])
                            right--;
                    }
                }
            }
        }
        return res;
    }
};

原文：http://www.cnblogs.com/Kai-Xing/p/4035330.html