问题描述

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.

问题分析

设置两个变量m1、m2记录当前最小值和次小值，在新读入一个数字x时：

• x<=m1：更新m1=x；
• m1<x<=m2：更新m2=x；
• 出现m1<m2<x的情况，返回True。

这个算法的巧妙之处在于保证m1<m2，并且m1、m2的更新很有趣，使当前保存的m1、m2始终最有利于获得解。

AC代码

class Solution(object):  
    def increasingTriplet(self, nums):  
        if len(nums) < 3:
            return False
        m1 = m2 = max(nums) + 1
        for num in nums:
            if num <= m1:
                m1 = num
            elif num <= m2:
                m2 = num
            else:
                return True
        return False

Runtime: 48 ms, which beats 81.85% of Python submissions.