所有Leetcode题目不定期汇总在 Github, 欢迎大家批评指正，讨论交流。
'''
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.
Your algorithm's runtime complexity must be in the order of O(log n).
Example 1:
Input: nums = [4,5,6,7,0,1,2], target = 0
Output: 4
Example 2:
Input: nums = [4,5,6,7,0,1,2], target = 3
Output: -1
'''
class Solution:
def search(self, nums: List[int], target: int) -> int:
if len(nums) <= 0 : return -1
l ,r = 0 , len(nums) - 1
while l <= r :
mid = l + (r - l) // 2
if nums[mid] == target: return mid
if nums[r] == target: return r
if nums[l] < nums[mid]:
if target >= nums[l] and nums[mid] >= target: r = mid - 1
else: l = mid + 1
else:
if target >= nums[mid] and target <= nums[r]: l = mid + 1
else: r = mid - 1
return -1
''''
Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand.
(i.e., [0,0,1,2,2,5,6] might become [2,5,6,0,0,1,2]).
You are given a target value to search. If found in the array return true, otherwise return false.
Example 1:
Input: nums = [2,5,6,0,0,1,2], target = 0
Output: true
Example 2:
Input: nums = [2,5,6,0,0,1,2], target = 3
Output: false
Follow up:
This is a follow up problem to Search in Rotated Sorted Array, where nums may contain duplicates.
Would this affect the run-time complexity? How and why?
'''
class Solution:
def search(self, nums: List[int], target: int) -> bool:
if len(nums) <= 0 : return False
l , r = 0 , len(nums) - 1
while l <= r :
mid = l + (r - l) // 2
if target == nums[l] or target == nums[r] or target == nums[mid] : return True
if nums[l] == nums[mid]: l += 1
elif nums[l] < nums[mid]:
if nums[l] < target < nums[mid]: r = mid -1
else: l = mid + 1
else:
if nums[mid] < target < nums[r]: l = mid + 1
else: r = mid - 1
return False
所有Leetcode题目不定期汇总在 Github, 欢迎大家批评指正，讨论交流。
