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。

 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     }

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