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。
1 public class Solution { 2 public int search(int[] A, int target) { 3 int len=A.length; 4 if(len==0) return -1; 5 return bsearch(A,target,0,len-1); 6 7 8 9 } 10 11 public int bsearch(int A[],int target,int low,int high) 12 { 13 if(low>high) return -1; 14 int idx=-1; 15 if(A[low]<=A[high]) //it must be YouXu,so binary search 16 17 { 18 while(low<=high) 19 { 20 int mid=(low+high)>>1; 21 if(A[mid]==target) 22 { 23 idx=mid; 24 return idx; 25 } 26 else if(A[mid]>target) 27 { 28 high=mid-1; 29 } 30 else low=mid+1; 31 } 32 33 } 34 else 35 { 36 int mid=(low+high)>>1; 37 if(A[mid]==target) 38 { 39 idx=mid; 40 return idx; 41 42 } 43 else 44 { 45 idx=bsearch(A,target,low,mid-1); 46 if(idx==-1) 47 { 48 idx=bsearch(A,target,mid+1,high); 49 50 } 51 52 53 54 55 } 56 57 58 59 60 61 62 } 63 64 return idx; 65 66 67 } 68 }