题目来源
https://leetcode.com/problems/search-in-rotated-sorted-array-ii/
Follow up for "Search in Rotated Sorted Array":
What if duplicates are allowed?
Would this affect the run-time complexity? How and why?
Write a function to determine if a given target is in the array.
题意分析
Input:
:type nums: List[int]
:type target: int
Output:
rtype: bool
Conditions:一个翻转的有序数组，数组元素可能重复，判断target是否在数组中
题目思路
关键就要区分边界，采用first表示下界，last表示上界，mid为中间点。如果mid为target，返回True；否则，判断mid，first，last是否相等，若相等则缩小搜索空间，之后判断target在哪个区间，判断条件为：1）target与nums[mid]的大小，target与nums[first]的大小。
AC代码（Python）
class Solution(object):
def search(self, nums, target):
"""
:type nums: List[int]
:type target: int
:rtype: bool
"""
size = len(nums)
first = 0
last = size - 1
while first <= last:
mid = (last + first) / 2
print(first,mid, last)
if nums[mid] == target:
return True
if nums[mid] == nums[first] == nums[last]:
first += 1; last -= 1
elif nums[first] <= nums[mid]:
if target < nums[mid] and target >= nums[first]:
last = mid - 1
else:
first = mid + 1
else:
if target >= nums[mid] and target < nums[first]:
first = mid + 1
else:
last = mid - 1 return False