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<=.

对数组nums排序，记排序后的数组为sorts，数组长度为n

foor loop 寻找当nums[i] != sorts[i]时的最小index和最大index

最后return r-l+1: 一般通过索引求数组长度就用左指针 - 右指针 + 1

class Solution(object):
    def findUnsortedSubarray(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """

        n, sorts = len(nums), sorted(nums)
        if nums == sorts: return 0
        l, r = min(i for i in range(n) if nums[i] != sorts[i]), max(i for i in range(n) if nums[i] != sorts[i])
        return r - l + 1