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.

在一个未排序的数组中查看是否存在三个元素，满足arr[i] < arr[j] < arr[k], (0 ≤ i < j < k ≤ n-1).
要求在线性时间找到，要求空间复杂度为O(1)。我们维护两个变量min1，min2, 使min1 <min2，并且min1在min2的左边，所以我们只要找到一个元素大于min2就可以返回true。这样时间复杂度为O(n)，空间复杂度为O(1)。代码如下：

public class Solution {
    public boolean increasingTriplet(int[] nums) {
        if(nums == null || nums.length < 3) return false;
        int min1 = Integer.MAX_VALUE;
        int min2 = Integer.MAX_VALUE;
        for(int i : nums) {
            if(i <= min1) min1 = i;
            else if(i <= min2) min2 = i;
            else if(i > min2) return true;
        }
        return false;
    }
}