linear-list/array/increasing-triplet-subsequence

Increasing Triplet Subsequence

描述

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

More specifically, if there exists i, j, k such that arr[i] < arr[j] < arr[k] given 0 ≤ i < j < k ≤ n-1 return true else return false.

Your function 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.

分析

扫描一遍数组，用变量x1保存当前最小的值，变量x2保存当前第二小的值。如果右面能碰到一个数大于x2，说明必然存在一个递增的三元组。

代码

// Increasing Triplet Subsequence
// Time complexity: O(n), Space complexity: O(1)
public class Solution {
    public boolean increasingTriplet(int[] nums) {
        int x1 = Integer.MAX_VALUE;
        int x2 = Integer.MAX_VALUE;

        for (int x : nums) {
            if (x <= x1) x1 = x;
            else if (x <= x2) x2 = x;
            else return true;
        }
        return false;
    }
}

相关题目

• Longest Increasing Subsequence