334. Increasing Triplet Subsequence

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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.

解法1： O(N)

类似于一个贪心的算法，我们可以设两个变量，这两个变量的条件就是a < b,同时a，b要尽可能的小，这样在后续的扫描中如果发现有一个数字比a，b都大我们就找到了答案。

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class Solution {
    public boolean increasingTriplet(int[] nums) {
        if (nums == null || nums.length < 3) {
            return false;
        }
        int small = Integer.MAX_VALUE, big = Integer.MAX_VALUE;
        for (int num : nums) {
            if (num <= small) small = num;
            else if (num <= big) big = num;
            else return true;
        }
        return false;
    }
}