【題目】
Follow up for "Search in RotatedSorted Array":
What if duplicates are allowed?
Would this affect the run-time complexity?How and why?
Suppose an array sorted in ascending orderis rotated at some pivot unknown to you beforehand.
(i.e., 0 1 2 4 5 6 7 might become 4 5 6 7 01 2).
Write a function to determine if a giventarget is in the array.
The array may contain duplicates.
對一個給定的有序數組，數組中存在重複數字，以某數字爲中心進行旋轉，在該數組中尋找目標target是否存在。
【思路】
方法一：直接查找
方法二：利用二分查找，因爲有旋轉的情況，需要注意邊界條件
【Python實現】
#二分查找
class Solution(object):
def search(self, nums, target):
"""
:type nums: List[int]
:type target: int
:rtype: bool
"""
left = 0
right = len(nums) - 1
while left <= right:
mid = (left + right)//2
if nums[mid] == target:
return True
if nums[left] == nums[right] == nums[mid]:
left += 1
right -= 1
elif nums[left] <= nums[mid]:#升序
if nums[left] <= target <nums[mid]:
right = mid -1
else:
left = mid + 1
else:#旋轉（左大右小）
if nums[left] > target >=nums[mid]:
left = mid + 1
else:
right = mid - 1
return False
if __name__ == '__main__':
nums = [3, 1]
target = 1
S = Solution()
S.search(nums,target)
#直接逐個查找
class Solution1(object):
def search(self, nums, target):
"""
:type nums: List[int]
:type target: int
:rtype: bool
"""
for i in nums:
if i == target:
return True
return False
if __name__ == '__main__':
nums= [3, 1]
target = 1
S = Solution1()
S.search(nums,target)
