leetcode 334. Increasing Triplet Subsequence

Given an unsorted array return whether an increasing subsequence of length 3 exists or not in the array.


Formally the function should:
Return true if there exists i, j, k 
such that arr[i] < arr[j] < arr[k] given 0 ≤ i < j < k ≤ n-1 else return false.
Your algorithm should run in O(n) time complexity and O(1) space complexity.


Examples:
Given [1, 2, 3, 4, 5],
return true.


Given [5, 4, 3, 2, 1],

return false.

#include "stdafx.h"
#include<vector>
#include<iostream>
using namespace std;

class Solution {
public:
	bool increasingTriplet(vector<int>& nums)
	{
		if (nums.size() < 3)
			return false;
		vector<vector<int>>gtthan;
		for (int i = 0; i < nums.size(); i++)
		{
			vector<int>temp;
			gtthan.push_back(temp);
			for (int j = i + 1; j < nums.size(); j++)
			{
				if (nums[j]>nums[i])
					gtthan[i].push_back(j);
			}
		}
		int num1, num2, num3;
		for (int i = 0; i < nums.size() - 2; i++)
		{
			num1 = nums[i];
			for (int j = 0; j < gtthan[i].size(); j++)
			{
				num2 = nums[gtthan[i][j]];
				for (int k = 0; k < gtthan[gtthan[i][j]].size(); k++)
				{
					//if (gtthan[gtthan[i][j]].size()>0)
					return true;
				}
			}
		}
		return false;
	}
};




int _tmain(int argc, _TCHAR* argv[])
{
	int a[5] = { 1, 2, -10, -8, -7};
	vector<int>nums(a, a + 5);
	Solution sl;
	cout<<sl.increasingTriplet(nums);
	system("pause");
	return 0;
}

acceped