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：

public class Solution {
    public boolean search(int[] A, int target) {
       int start=0,end=A.length-1,mid=0;
        while(start<=end){
            mid=(start+end)/2;
            if(A[mid]==target)return true;
            if(A[start]<A[mid]){
                if((target<A[mid])&&(target>=A[start]))
                    end=mid-1;
                    else start=mid+1;
            }
            else if(A[start]>A[mid]){
                if((target<A[mid])||(target>=A[start]))
                    end=mid-1;
                    else start=mid+1;
            }
            else start++;
        }
        return false;
    }
}

python：

class Solution:
    # @param A, a list of integers
    # @param target, an integer to be searched
    # @return an integer

    def search(self, A, target):
        left = 0
        right = len(A) - 1
        while(left <= right):
            mid = (left + right) / 2
            if (A[mid] == target):
                return True
            if (A[left] < A[mid]):
                if (target >= A[left] and target < A[mid]):
                    right = mid - 1
                else:
                    left = mid + 1
            elif (A[left] > A[mid]):
                if (target > A[mid] and target <= A[right]):
                    left = mid + 1
                else:
                    right = mid - 1
            else:
                left += 1
        return False