[LeetCode] 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.

解题思路

初始化时设置n1、n2为0x7fffffff。遍历数组：

• 若n小于等于n1，则n1=n;
• 否则，若n小于等于n2，则n2=n;
• 否则，返回true;

该过程不断缩小n1、n2，最后查看是否具有比n1、n2都小的数。

实现代码

// Runtime: 1 ms
public class Solution {public boolean increasingTriplet(int[] nums) {int n1 = 0x7fffffff;int n2 = 0x7fffffff;for (int n : nums) {if (n <= n1) {n1 = n;} else if (n <= n2) {n2 = n;} else {return true;}}return false;}
}

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