题目：
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.
题解：
这道题与之前Search in Rotated Sorted Array类似，问题只在于存在dupilcate。那么和之前那道题的解法区别就是，不能通过比较A[mid]和边缘值来确定哪边是有序的，会出现A[mid]与边缘值相等的状态。所以，解决方法就是对于A[mid]==A[low]和A[mid]==A[high]单独处理。
当中间值与边缘值相等时，让指向边缘值的指针分别往前移动，忽略掉这个相同点，再用之前的方法判断即可。
这一改变增加了时间复杂度，试想一个数组有同一数字组成{1，1，1，1，1}，target=2, 那么这个算法就会将整个数组遍历，时间复杂度由O(logn)升到O(n)
实现代码如下：
1 public boolean search(int [] A,int target){
2 if(A==null||A.length==0)
3 return false;
4
5 int low = 0;
6 int high = A.length-1;
7
8 while(low <= high){
9 int mid = (low + high)/2;
10 if(target < A[mid]){
11 if(A[mid]<A[high])//right side is sorted
12 high = mid - 1;//target must in left side
13 else if(A[mid]==A[high])//cannot tell right is sorted, move pointer high
14 high--;
15 else//left side is sorted
16 if(target<A[low])
17 low = mid + 1;
18 else
19 high = mid - 1;
20 }else if(target > A[mid]){
21 if(A[low]<A[mid])//left side is sorted
22 low = mid + 1;//target must in right side
23 else if(A[low]==A[mid])//cannot tell left is sorted, move pointer low
24 low++;
25 else//right side is sorted
26 if(target>A[high])
27 high = mid - 1;
28 else
29 low = mid + 1;
30 }else
31 return true;
32 }
33
34 return false;
35 }
2 if(A==null||A.length==0)
3 return false;
4
5 int low = 0;
6 int high = A.length-1;
7
8 while(low <= high){
9 int mid = (low + high)/2;
10 if(target < A[mid]){
11 if(A[mid]<A[high])//right side is sorted
12 high = mid - 1;//target must in left side
13 else if(A[mid]==A[high])//cannot tell right is sorted, move pointer high
14 high--;
15 else//left side is sorted
16 if(target<A[low])
17 low = mid + 1;
18 else
19 high = mid - 1;
20 }else if(target > A[mid]){
21 if(A[low]<A[mid])//left side is sorted
22 low = mid + 1;//target must in right side
23 else if(A[low]==A[mid])//cannot tell left is sorted, move pointer low
24 low++;
25 else//right side is sorted
26 if(target>A[high])
27 high = mid - 1;
28 else
29 low = mid + 1;
30 }else
31 return true;
32 }
33
34 return false;
35 }