334. Increasing Triplet Subsequence【M】【69】_increasingtriplet-优快云博客

本文介绍了一种高效算法，用于判断一个未排序数组中是否存在长度为3的递增子序列。该算法的时间复杂度为O(n)，空间复杂度为O(1)。

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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.

Credits:
Special thanks to @DjangoUnchained for adding this problem and creating all test cases.

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就是维护一个大小为3的窗口

low代表最小的，mid代表中间大小的

如果当前数，大于mid，那肯定满足条件

这里面有个trick，就是low的位置不一定在mid前面，它可以代表有可能成为最小的的候选。

class Solution(object):
    def increasingTriplet(self, nums):
        if len(nums)<3:
            return False

        mid=sys.maxint
        low=nums[0]
        for i in nums[1:]:
            #print low,mid
            if i > mid:
                return True
            if i > low:
                mid = min(mid,i)
            low = min(low,i)
        return False