【python3】leetcode 581. Shortest Unsorted Continuous Subarray（easy）

最新推荐文章于 2021-08-03 15:30:54 发布

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最新推荐文章于 2021-08-03 15:30:54 发布

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分类专栏： leetcode python

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本文介绍了一种寻找最短未排序连续子数组的方法，通过先对数组进行排序，然后找出首尾不匹配的位置，以此确定需排序的子数组范围。提供了两种实现方式，一种使用两个while循环，另一种则在一个while循环中完成，后者效率较低。

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581. Shortest Unsorted Continuous Subarray（easy）

Given an integer array, you need to find one continuous subarray that if you only sort this subarray in ascending order, then the whole array will be sorted in ascending order, too.

You need to find the shortest such subarray and output its length.

Example 1:
Input: [2, 6, 4, 8, 10, 9, 15]
Output: 5
Explanation: You need to sort [6, 4, 8, 10, 9] in ascending order to make the whole array sorted in ascending order.

哇这个真的是最近写的最长的代码了不过逻辑很简单就是先把nums排序

然后找第一个和最后一个不相等的分别为未排序子串的首尾start 和end,

用两个while比用一个while省时间因为不用遍历完整个list

一、两个while

class Solution:
    def findUnsortedSubarray(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        length = len(nums)
        sortnums = sorted(nums)
        if length == 1 or sorted(nums) == nums:return 0
        start = 0
        end = length - 1
        i = 0
        while(i < length):
            if nums[i] != sortnums[i]:
                start = i
                break
            i += 1
        i = length - 1
        while(i >= 0):
            if nums[i] != sortnums[i]:
                end = i
                break
            i -= 1
        return end - start + 1

Runtime: 96 ms, faster than 66.33% of Python3

二、一个while

class Solution:
    def findUnsortedSubarray(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        length = len(nums)
        sortnums = sorted(nums)
        if length == 1 or sorted(nums) == nums:return 0
        start = length
        end = 0
        i = 0
        while(i < length):
            if nums[i] != sortnums[i]:
                start = min(start,i)
                end = max(end,i)
            i += 1
        return end - start + 1

Runtime: 128 ms, faster than 37.42% of Python3