[LeetCode] 81. Search in Rotated Sorted Array II_Medium tag: not real binary search anymore
2019-07-12 10:53 Johnson_强生仔仔 阅读(161) 评论(0) 编辑 收藏 举报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?
这个题目跟[LeetCode] 154. Find Minimum in Rotated Sorted Array II_Hard tag: not real binary search anymore类似,因为可以有重复,如果重复数目太多的话,时间复杂度会上升到O(n),所以就直接用O(n) 的方法就好。
Code
class Solution: def searchRotate2(self, nums, target): return True if target in nums else False
Update on 08/01/2023. 看edge case, 1,0,1,1,1,1,1 , 这个情况那就是nums[l] == nums[mid] == nums[r], 就一直r -= 1 and l += 1 直到他们不相等。然后再二分
Code:
class Solution: def search(self, nums: List[int], target: int) -> bool: l, r = 0, len(nums) - 1 while l + 1 < r: mid = l + (r - l)//2 num = nums[mid] if num == target: return True # for edge case 3, 1, 2, 3, 3, 3, 3, 3 # for edge case 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, find the mininal first index, but does not mean that index to n - 1 are not decreasing order. elif nums[l] == num == nums[r]: l += 1 r -= 1 # first half elif nums[l] <= num: if nums[l] <= target <= num: r = mid else: l = mid # second half else: if num <= target <= nums[r]: l = mid else: r = mid return True if target in [nums[l],nums[r]] else False