# Topic 1

Website: 15. The sum of three numbers.

Title Description:

Given a n array nums containing n integers, determine whether there are three elements a, b, c in nums, so that a + b + c = 0? Find out all triples that satisfy the conditions and do not repeat.

Note: The answer should not contain duplicate triples.

Example 1:

```For example, given array nums = [-1, 0, 1, 2, -1, -4],

The set of triples satisfying the requirements is:
[
[-1, 0, 1],
[-1, -1, 2]
]

Source: LeetCode
```

# Two, procedure

```#include <iostream>
#include <vector>
#include <algorithm>

using namespace std;

//Sort first, then double pointer method, 160 ms, memory consumption 18MB
vector<vector<int> > threeSum(vector<int>& nums)
{

vector<vector<int> > res;
if(nums.size()<3) return res;
sort(nums.begin(),nums.end());
for(int i=0;i<nums.size()-2;i++)
{
int j=i+1;
int k=nums.size()-1;
while (j<k)
{
if(nums[i]>0) break;
if (nums[i]+nums[j]+nums[k]<0)
{
j++;
}
else if (nums[i]+nums[j]+nums[k]>0)
{
k--;
}
else
{
vector<int> temp;
temp.push_back(nums[i]);
temp.push_back(nums[j]);
temp.push_back(nums[k]);
res.push_back(temp);
while(j<nums.size()-2 && nums[j]==nums[j+1]) j++;
j++;
}
}
while(i<nums.size()-3 && nums[i]==nums[i+1]) i++;
}

return res;
}

int main()
{
int a[]={-1,0,1,2,-1,-4};
vector<int> nums(a,a+sizeof(a)/sizeof(int));
vector<vector<int>> res;
res=threeSum(nums);
for (int i=0;i<res.size();i++)
{
for (int j=0;j<res[0].size();j++)
{
printf("%d ",res[i][j]);
}
printf("\n");
}

system("pause");
return -1;
}
```

Result:

Keywords: network

Added by xposed on Wed, 02 Oct 2019 21:07:11 +0300