http://blog.csdn.net/pickless/article/details/9191075
Suppose a sorted array is rotated at some pivot unknown to you beforehand.
(i.e., 0 1 2 4 5 6 7 might become 4 5 6 7 0 1 2).
You are given a target value to search. If found in the array return its index, otherwise return -1.
You may assume no duplicate exists in the array.
设置 low=0 high=len-1
bsearch(A,low,high)=bsearch(A,low,mid-1)||bsearch(A,mid+1,high) (A[mid]!=target)
mid (A[mid]==target)
这是传统的思路：
我们经过观察发现，只要 A[high]>=A[low] 数组一定是递增的，则选用二分查找
如果不是，则分割 ，使用递归，一定能分割成数组满足第一种状态，
比如，上边的例子我们查找 5，
mid=3,A[mid]=7；分割成（4,5,6）和（0,1,2）连个尾巴都是大于首部，所以直接会被二分查找。
public class Solution {
public int search(int[] A, int target) {
int len=A.length;
if(len==0) return -1;
return bsearch(A,target,0,len-1);
}
public int bsearch(int A[],int target,int low,int high)
{
if(low>high) return -1;
int idx=-1;
if(A[low]<=A[high]) //it must be YouXu,so binary search
{
while(low<=high)
{
int mid=(low+high)>>1;
if(A[mid]==target)
{
idx=mid;
return idx;
}
else if(A[mid]>target)
{
high=mid-1;
}
else low=mid+1;
}
}
else
{
int mid=(low+high)>>1;
if(A[mid]==target)
{
idx=mid;
return idx;
}
else
{
idx=bsearch(A,target,low,mid-1);
if(idx==-1)
{
idx=bsearch(A,target,mid+1,high);
}
}
}
return idx;
}
}