leetcode旋转数组查找 二分查找的变形

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;

     }
     }