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==0||n==1&&A[0]!=target) return -1;
        if(A[0]==target) return 0;
        int i=1;
        while(A[i-1]<A[i])  i++;
        
        int pre=binary_search(A,0,i,target);
        int pos=binary_search(A,i,n-i,target);
        return pre==-1?pos:pre;
    }
private:
    int binary_search(int *B,int lo,int len,int goal)
    {
        int low=lo;
        int high=lo+len-1;
        while(low<=high)
        {
            int middle=(low+high)/2;
            if(goal==B[middle])//找到，返回index
               return middle;
            else if(B[middle]<goal)//在右边
               low=middle+1;
            else//在左边
               high=middle-1;
        }
        return -1;//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 = 0, last = n-1;
        while (first <= last) 
        {
            const int mid = (first + last) / 2;
            if (A[mid] == target)
                return mid;
            if (A[first] <= A[mid]) 
            {
                if (A[first] <= target && target < A[mid])
                   last = mid-1;
                else
                   first = mid + 1;
            } 
            else 
            {
                if (A[mid] < target && target <= A[last])
                    first = mid + 1;
                else
                    last = mid-1;
            }
        }
        return -1;
    }
};