【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;

}