leetcode------Search in Rotated Sorted ArrayII

题目	Search in Rotated Sorted ArrayII
通过率：	31.4%
难度：	中等

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.

本题意思是会出现重复元素

如:456701234;

但是我发现我再第一版本中的做法是不受本题影响的，具体参照我第一个版本的做法

代码如下：

 1 public class Solution {

 2     public boolean search(int[] A, int target) {

 3        int start=0,end=A.length-1,mid=0;

 4         while(start<=end){

 5             mid=(start+end)/2;

 6             if(A[mid]==target)return true;

 7             if(A[start]<A[mid]){

 8                 if((target<A[mid])&&(target>=A[start]))

 9                     end=mid-1;

10                     else start=mid+1;

11             }

12             else if(A[start]>A[mid]){

13                 if((target<A[mid])||(target>=A[start]))

14                     end=mid-1;

15                     else start=mid+1;

16             }

17             else start++;

18         }

19         return false;

20     }

21 }

python：

 1 class Solution:

 2     # @param A, a list of integers

 3     # @param target, an integer to be searched

 4     # @return an integer

 5 

 6     def search(self, A, target):

 7         left = 0

 8         right = len(A) - 1

 9         while(left <= right):

10             mid = (left + right) / 2

11             if (A[mid] == target):

12                 return True

13             if (A[left] < A[mid]):

14                 if (target >= A[left] and target < A[mid]):

15                     right = mid - 1

16                 else:

17                     left = mid + 1

18             elif (A[left] > A[mid]):

19                 if (target > A[mid] and target <= A[right]):

20                     left = mid + 1

21                 else:

22                     right = mid - 1

23             else:

24                 left += 1

25         return False