【leetcode】3Sum (medium)

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.

 

思路：

把最小值从头到尾循环，中间值和最大值两边收缩。

我写的时候是在最后去重复，总是超时。后来看了人家的答案，发现可以每次对最小、中间、最大值去重。就可以AC了

#include <iostream>
#include <vector>
#include <algorithm>
#include <queue>
#include <stack>
using namespace std;

class Solution {
public:
    vector<vector<int> > threeSum(vector<int> &num) {
        vector<vector<int> > ans;
        if(num.size() < 3)
        {
            return ans;
        }
        int small = 0;
        int big = num.size() - 1;
        int mid = 0;
        int sum = 0;
        sort(num.begin(), num.end());
        for(small = 0; small < num.size() - 2; /*small++*/) //最小数字循环  中间与最大数字两边收缩
        {
            mid = small + 1;
            big = num.size() - 1;
            while(mid < big)
            {
                sum = num[small] + num[mid] + num[big];
                if(sum > 0)
                {
                    big--;
                }
                else if(sum < 0)
                {
                    mid++;
                }
                else
                {
                    vector<int> v;
                    v.push_back(num[small]);
                    v.push_back(num[mid]);
                    v.push_back(num[big]);
                    ans.push_back(v);
                    do { mid++; }while (mid < big && num[mid - 1] == num[mid]); //注意！！
                    do { big--; }while (mid < big && num[big + 1] == num[big]);//注意！！
                }
            }
            do{ small++; }while (small < num.size() - 1 && num[small - 1] == num[small]);//注意！！
        }
        return ans;
    }
};

int main()
{
    Solution s;
    vector<int> num;
    num.push_back(-4);
    num.push_back(-1);
    num.push_back(-1);
    num.push_back(-1);
    num.push_back(-1);
    num.push_back(0);
    num.push_back(0);
    num.push_back(0);
    num.push_back(1);
    num.push_back(2);

    vector<vector<int>> ans = s.threeSum(num);

    return 0;


}