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Leetcode 978. Longest Turbulent

Leetcode 978. Longest Turbulent

作者: SnailTyan | 来源:发表于2021-08-19 08:59 被阅读0次

    文章作者:Tyan
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    1. Description

    Longest Turbulent Subarray

    2. Solution

    解析:Version 1,根据题意,数字的大小关系一直在反转,因此这里采用一个布尔值来表示下一个比较状态,每次比较后都将布尔值反转,只要比较状态和布尔值相等,则动荡序列的长度加1。由于初始布尔值状态未知,因此设为None。当前数字与下一个数字相等时,布尔值设为None,计数器设为1。依次比较数字时,首先排除数字相等的状态,前后两个数字相等时,以下一个数字作为初始序列数字,重新计数。当数字比较顺序与布尔值不相等时,则对当前两个数字的比较状态取反作为下一次比较的状态,且当前两个数字应该作为初始序列长度,因此count=2,布尔值不更新。Version 2是动态规划,Version 3是另一种形式的动态规划。

    • Version 1
    class Solution:
        def maxTurbulenceSize(self, arr: List[int]) -> int:
            n = len(arr)
            maximum = 1
            count = 0
            flag = None
            for i in range(n-1):
                if arr[i] == arr[i+1]:
                    count = 1
                    flag = None
                    continue
                if flag is None:
                    flag = not (arr[i] > arr[i+1])
                    count = 2
                else:
                    if (arr[i] > arr[i+1]) == flag:
                        count += 1
                        flag = not flag
                    else:
                        # flag = not (arr[i] > arr[i+1])
                        count = 2
                maximum = max(maximum, count)
            return maximum
                
    
    • Version 2
    class Solution:
        def maxTurbulenceSize(self, arr: List[int]) -> int:
            n = len(arr)
            high = [1] * n
            low = [1] * n
            for i in range(n-1):
                if arr[i] > arr[i+1]:
                    high[i+1] = low[i] + 1
                elif arr[i] < arr[i+1]:
                    low[i+1] = high[i] + 1
            maximum = max(max(high), max(low))
            return maximum
    
    • Version 3
    class Solution:
        def maxTurbulenceSize(self, arr: List[int]) -> int:
            n = len(arr)
            maximum = 1
            high = 1
            low = 1
            for i in range(n-1):
                if arr[i] > arr[i+1]:
                    high = low + 1
                    low = 1
                elif arr[i] < arr[i+1]:
                    low = high + 1
                    high = 1
                else:
                    high = 1
                    low = 1
                maximum = max(maximum, max(high, low))
            return maximum
    

    Reference

    1. https://leetcode.com/problems/longest-turbulent-subarray/

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