leetcode题解：Search in Rotated Sorted Array（旋转排序数组查找）

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.

说明：

1）已排序数组查找采用二分查找

2）关键找到临界点

实现：

一、我的代码：

     

 class Solution {
 public:
     int search(int A[], int n, int target) {
         if(n==||n==&&A[]!=target) return -;
         if(A[]==target) return ;
         int i=;
         while(A[i-]<A[i])  i++;

         int pre=binary_search(A,,i,target);
         int pos=binary_search(A,i,n-i,target);
         return pre==-?pos:pre;
     }
 private:
     int binary_search(int *B,int lo,int len,int goal)
     {
         int low=lo;
         int high=lo+len-;
         while(low<=high)
         {
             int middle=(low+high)/;
             if(goal==B[middle])//找到，返回index
                return middle;
             else if(B[middle]<goal)//在右边
                low=middle+;
             else//在左边
                high=middle-;
         }
         return -;//没有，返回-1
     }
 };

二、网上开源代码：

 class Solution {
 public:
     int search(int A[], int n, int target) {
         int first = , last = n-;
         while (first <= last)
         {
             const int mid = (first + last) / ;
             if (A[mid] == target)
                 return mid;
             if (A[first] <= A[mid])
             {
                 if (A[first] <= target && target < A[mid])
                    last = mid-;
                 else
                    first = mid + ;
             }
             else
             {
                 if (A[mid] < target && target <= A[last])
                     first = mid + ;
                 else
                     last = mid-;
             }
         }
         return -;
     }
 };