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82 lines (74 loc) · 1.21 KB
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/**
* 3Sum
* Two pointer, time complexity is O(n^2))
*
* cpselvis(cpselvis@gmail.com)
* September 17th, 2016
*/
#include<iostream>
#include<vector>
using namespace std;
class Solution {
public:
vector<vector<int> > threeSum(vector<int>& nums) {
vector<vector<int> > ret;
int n = nums.size();
if (n < 3)
{
return ret;
}
sort(nums.begin(), nums.end());
for (int i = 0; i < n; i ++)
{
// Avoid repeate situation.
if (i > 0 && nums[i] == nums[i - 1])
{
continue;
}
int l = i + 1, r = n - 1;
while (l < r)
{
int sum = nums[i] + nums[l] + nums[r];
if (sum == 0)
{
ret.push_back(vector<int>{nums[i], nums[l], nums[r]});
// Avoid repeate elements.
while (nums[l] == nums[l + 1])
{
l ++;
}
while (nums[r] == nums[r - 1])
{
r --;
}
l ++;
r --;
}
else if (sum > 0)
{
r --;
}
else
{
l ++;
}
}
}
return ret;
}
};
int main(int argc, char **argv)
{
int arr[6] = {-1, 0, 1, 2, -1, -4};
vector<int> vec(arr + 0, arr + 6);
Solution s;
vector<vector<int> > ret = s.threeSum(vec);
for (auto i : ret)
{
for (auto j : i)
{
cout << j << " ";
}
cout << endl;
}
}