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:
- Then length of the input array is in range [1, 10,000].
- The input array may contain duplicates, so ascending order here means <=.
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刚开始想从左到右看如果一直递增,忽然到i位置开始drop,那么这个drop的位置到底是历史第几低呢???感觉又要二分了,所以就没往下想。。。跑去排序了
但是这个题目O(n)的解法恰好利用了这点:从左到右看如果一直递增,忽然到i位置开始drop,那么包括i在内i之前的肯定要排序,虽然起点并不知道;同理,从右到左看如果一直递减,忽然到j位置开始increase,那么包括j在内的之后的肯定要排序。通过这种方式可以确定[j,i]
另外,这个题的起点和终点要遍历整个数组才知道,所以注意边界条件。
class Solution:
def findUnsortedSubarray(self, nums: List[int]) -> int:
l = len(nums)
if (l <= 1):
return 0
max_from_left,min_from_right,beg,end=nums[0],nums[l-1],-1,-2
for i in range(1,l):
max_from_left = max(max_from_left,nums[i])
min_from_right = min(min_from_right,nums[l-1-i])
if (nums[i] < max_from_left):
end = i
if (nums[l-1-i] > min_from_right):
beg = l-1-i
return 0 if beg>end else end-beg+1