LeetCode81:Search in Rotated Sorted Array II
阿新 • • 發佈:2018-11-29
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?
LeetCode:連結
在LeetCode33:Search in Rotated Sorted Array
當 nums[mid] = nums[left] 時,這時由於很難判斷 target 會落在哪,那麼只能採取 left++
當 nums[mid] > nums[left] 時,這時可以分為兩種情況,判斷左半部比較簡單
當 nums[mid] < nums[left] 時,這時可以分為兩種情況,判斷右半部比較簡單
class Solution(object): def search(self, nums, target): """ :type nums: List[int] :type target: int :rtype: int """ low = 0 high = len(nums) - 1 while low <= high: mid = (low + high) // 2 if nums[mid] == target: return True if nums[mid] == nums[low]: low += 1 elif nums[mid] > nums[low]: if target >= nums[low] and target < nums[mid]: high = mid - 1 else: low = mid + 1 else: if target > nums[mid] and target <= nums[high]: low = mid + 1 else: high = mid - 1 return False