Search in Rotated Sorted Array leetcode java

题目：

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.

题解：

这道题是一道常见的二分查找法的变体题。

要解决这道题，需要明确rotated sorted array的特性，那么就是至少有一侧是排好序的（无论pivot在哪，自己画看看）。接下来就只需要按照这个特性继续写下去就好。以下就以伪代码方法来说明：

1. 如果target比A[mid]值要小
2.       如果A[mid]右边有序（A[mid]<A[high])
   
3.             那么target肯定不在右边（target比右边的都得小），在左边找
4.       如果A[mid]左边有序
5.             那么比较target和A[low]，如果target比A[low]还要小，证明target不在这一区，去右边找；反之，左边找。
6. 如果target比A[mid]值要大
7.      如果A[mid]左边有序（A[mid]>A[low]）
   
8.            那么target肯定不在左边（target比左边的都得大），在右边找
   
9.      如果A[mid]右边有序
10.            那么比较target和A[high]，如果target比A[high]还要大，证明target不在这一区，去左边找；反之，右边找。

以上实现代码如下所示：

 1    public int search(int [] A,int target){
 2        if(A==null||A.length==0)
 3          return -1;
 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
14                  if(target<A[low])//target<A[mid]&&target<A[low]==>means,target cannot be in [low,mid] since this side is sorted
15                     low = mid + 1;
16                  else 
17                     high = mid - 1;
18            }else if(target > A[mid]){
19                if(A[low]<A[mid])//left side is sorted
20                  low = mid + 1;//target must in right side
21                else
22                  if(target>A[high])//right side is sorted. If target>A[high] means target is not in this side
23                     high = mid - 1;
24                  else
25                     low = mid + 1;
26            }else
27              return mid;
28        }
29        
30        return -1;
31 }

 Reference:http://www.cnblogs.com/ider/archive/2012/04/01/binary_search.html