## Description

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 <=.

## Solutions

### 1. 排序

直接排序然后比较不同，结果 TLE 了。

class Solution:
def findUnsortedSubarray(self, nums: List[int]) -> int:
copy = [num for num in nums]
n = len(nums)
self.quick_sort(nums, 0, n-1)
min_i, max_i = 0, 0
for i in range(n):
if nums[i] != copy[i]:
min_i = i
break

for i in range(n-1, -1, -1):
if nums[i] != copy[i]:
max_i = i
break
if max_i == min_i:
return 0
return max_i - min_i + 1

def quick_sort(self, nums, l=0, r=None):
if l > r:
return
pivot = self.partition(nums, l, r)
self.quick_sort(nums, l, pivot-1)
self.quick_sort(nums, pivot+1, r)

def partition(self, nums, l, r):
pivot = nums[r]
i = l
for j in range(l, r+1):
if nums[j] < pivot:
nums[i], nums[j] = nums[j], nums[i]
i += 1
nums[i], nums[r] = nums[r], nums[i]
return i


### 2. 最大最小扫描

# Time: O(n)
# Space: O(1)
class Solution:
def findUnsortedSubarray(self, nums: List[int]) -> int:
if not nums:
return 0

n = len(nums)
max_v, min_v = float('-inf'), float('inf')
p, q = 0, 0
for i in range(n):
if nums[i] >= max_v:
max_v = nums[i]
else:
p = i
for i in range(n-1, -1, -1):
if nums[i] <= min_v:
min_v = nums[i]
else:
q = i

if p == q:
return 0
return p - q + 1
# 307/307 cases passed (216 ms)
# Your runtime beats 87.91 % of python3 submissions
# Your memory usage beats 100 % of python3 submissions (13.7 MB)