LeetCode：(Array-33)Search in Rotated Sorted Array

去哪网面试题:二分查找的变形题目

　　目的是为了在O(logn)的时间复杂度下解决此问题，所以用二分查找。

package com.hb.leetcode;


/*
 * Search in Rotated Sorted  Array
 * 
 * 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.
 * 
 */

public class SearchInRotated {
    
    public  static int search(int[] A , int  target){
        return rec( A , target , 0 , A.length -1 );
    }
     
    public  static int rec(int[]  A , int target , int  low , int  high){
        if(low > high){  //没找到的情况
            return -1 ;  
        } 
        
        int  mid = low + (high - low) / 2 ;
        
        if(A[mid] == target){  //找到了
            return mid;
        }
        
        int res = rec(A , target , low , mid -1 );  //在左侧查找 
        
        if(res == -1 ){  //如果左侧没有找到，继续在右侧查找
            res = rec(A , target , mid + 1 , high ); 
        }
        
        
        return res;
    }
    
    public static void main(String[] args) {
        int[]  A = { 2, 3, 4, 5, 6, 0 ,1};
        System.out.println(search(A, 0));
    }
    
    

}