LeetCode-Search in Rotated Sorted Array

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.

思路分析:

(1)如果left<right,直接用二分搜索

(2)if left>right

     则 mid=(left+right)>>1;

　　if (A[mid])==target 则停止

　　else find(A,left,mid-1,target)与find(A,mid+1,right,target).

源代码:

class Solution {
public:
    int search(int A[], int n, int target) {
        return find(A,0,n-1,target);
    }
    
private:
   int find(int A[],int left,int right,int target){
       if(left>right) return -1;
       
       int idx=-1;
       
       if(A[left]<=A[right]){
           while(left<=right){
               int mid=(left+right)>>1;
               
               if(A[mid]==target){
                   idx=mid;
                   break;
               }
               else if(A[mid]>target) right=mid-1;
               else left=mid+1;
           }
       }
       
       else{
           int mid=(left+right)>>1;
           if(A[mid]==target) idx=mid;
           else{
               idx=find(A,left,mid-1,target);
               idx=((idx==-1)?find(A,mid+1,right,target):idx);
           }
       }
       return idx;
   }
};