Search in Rotated Sorted Array leetcode java

題目：

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.

 

題解：

這道題是一道常見的二分查詢法的變體題。

要解決這道題，需要明確rotated sorted array的特性，那麼就是至少有一側是排好序的（無論pivot在哪，自己畫看看）。接下來就只需要按照這個特性繼續寫下去就好。以下就以虛擬碼方法來說明：

1. 如果target比A[mid]值要小
2.       如果A[mid]右邊有序（A[mid]<A[high])
   
3.             那麼target肯定不在右邊（target比右邊的都得小），在左邊找
4.       如果A[mid]左邊有序
5.             那麼比較target和A[low]，如果target比A[low]還要小，證明target不在這一區，去右邊找；反之，左邊找。
6. 如果target比A[mid]值要大
7.      如果A[mid]左邊有序（A[mid]>A[low]）
   
8.            那麼target肯定不在左邊（target比左邊的都得大），在右邊找
   
9.      如果A[mid]右邊有序
10.            那麼比較target和A[high]，如果target比A[high]還要大，證明target不在這一區，去左邊找；反之，右邊找。

以上實現程式碼如下所示：

 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