581. Shortest unsorted continuous Subarray shortest unsorted continuous sub-array

Given an integer array, need to find one continuous subarray So if you are only sort this subarray in ascending Order, then the whole array would 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: 5Explanation: need to sort [6, 4, 8, ten, 9] in ascending order to make the whole array sorte D in ascending order.

Note:

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

  
 

   
  

  
class Solution:
   
  

  
 def findUnsortedSubarray(self, nums):
   
  

  
 """
   
  

  
 :type nums: List[int]
   
  

  
 :rtype: int
   
  

  
 """
   
  

  
 copyNums = nums[::]
   
  

  
 copyNums.sort()
   
  

  
 left = len(nums) - 1
   
  

  
 right = len(nums) - 1
   
  

  
 for i in range(0, len(nums)):
   
  

  
 if nums[i] != copyNums[i]:
   
  

  
 left = i
   
  

  
 break
   
  

  

   
  

  
  for   in   range   (  Len   ( nums    -   1  ,  Span class= "PLN" >  0     - 1    
   
  

  
 if nums[i] != copyNums[i]:
   
  

  
 right = i
   
  

  
 break
   
  

  

   
  

  
 if right - left == 0:
   
  

  
 return 0
   
  

  
 else:
   
  

  
 return (right - left) + 1
  
 

From for notes (Wiz)

581. Shortest unsorted continuous Subarray shortest unsorted continuous sub-array