shortest-unsorted-continuous-subarray

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.

 

Note:

1. Then length of the input array is in range [1, 10,000].
2. The input array may contain duplicates, so ascending order here means <=.

 

First sort the original sequence , Then compare with the original sequence , Look at the sequence before and after which start is different from the sorted sequence , And then subtract ：

For example, in the above example

2, 6, 4, 8, 10, 9, 15

After sorting

2,4,6,8,9,10,15

So start from 4 The beginning is the first 2 A different , It starts later 10 Different , That is the first. 6 position , therefore 6-2+1=5

therefore ：

class Solution:
    def findUnsortedSubarray(self, nums: List[int]) -> int:
        nums2=sorted(nums)
        right=0
        left=-1
        for index in range(0,len(nums)):
            if nums[index]!=nums2[index]:
                left=index
        for index in range(len(nums)-1,-1,-1):
            if nums[index]!=nums2[index]:
                right=index
        return left-right+1

Because I didn't know before sorted This method can return the sorted sequence , So two deep copies are used (copy.deepcopy) Resulting in slow running speed .