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;

   }

};