- 033 search in rotated sorted arr
- 33 & 81. Search in Rotated S
- leetcode:数组(未完待续)
- 33. Search in Rotated Sorted Arr
- 33. Search in Rotated Sorted Arr
- 33. Search in Rotated Sorted Arr
- 81. Search in Rotated Sorted Arr
- 33. Search in Rotated Sorted Arr
- 33. Search in Rotated Sorted Arr
- 81. Search in Rotated Sorted Arr
search in rotated sorted array
Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand.
(i.e., 0 1 2 4 5 6 7 might become 4 5 6 7 0 1 2).
You are given a target value to search. If found in the array return its index, otherwise return -1.
You may assume no duplicate exists in the array.
不要用语言自带的库来解决,这个就是用二分法实现的,还是来个实在的:
class Solution:
def search(self, nums, target):
"""
:type nums: List[int]
:type target: int
:rtype: int
"""
lo, hi = 0, len(nums) - 1
while lo < hi:
mid = (lo + hi) // 2
if (nums[0] > target) ^ (nums[0] > nums[mid]) ^ (target > nums[mid]):
lo = mid + 1
else:
hi = mid
return lo if target in nums[lo:lo+1] else -1
这个异或用的实在是屌。。
再来个好理解的:
class Solution:
def search(self, nums, target):
"""
:type nums: List[int]
:type target: int
:rtype: int
"""
if not nums:
return -1
low, high = 0, len(nums) - 1
while low <= high:
mid = low + (high - low) // 2
if target == nums[mid]:
return mid
if nums[low] <= nums[mid]:
if nums[low] <= target <= nums[mid]:
high = mid - 1
else:
low = mid + 1
else:
if nums[mid] <= target <= nums[high]:
low = mid + 1
else:
high = mid - 1
return -1
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