[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:



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

.

Solution:

这里只需求出BIS3即可。

1、找到最小的数min1, 其位置为k1.

2、找到第二小的数min2, 其位置为k2 > k1.

3、如果k3（k3>k2）位置的数n>min2, 找到递增三元组(min1, min2, n)

　　如果n<=min1, 更新min1=n和k1 = k3, 如果n<=min2, 更新k2和min2.

bool increasingTriplet(vector<int>& nums)
{
int min1 = INT_MAX, min2 = INT_MAX;

for (int i = 0; i < nums.size(); i++)
{
if (nums[i] <= min1)
min1 = nums[i];
else if (nums[i] <= min2)
min2 = nums[i];
else
return true;
}

return false;
}