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 * Two pointer, time complexity is O(n^2))
 * cpselvis(cpselvis@gmail.com)
 * September 17th, 2016
#include<iostream>
#include<vector>
using namespace std;
class Solution {
  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])
      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])
	  while (nums[r] == nums[r - 1])
	else if (sum > 0)
    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;

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