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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.

//vs2012测试代码
#include<iostream>

using namespace std;

#define N 5

//observation is the key, try to solve it by modifying binary search
class Solution {
public:
    int search(int A[], int n, int target)
	{
		if( n==0 )
			return -1;
        int mid = 0;
		int left=0, right=n-1;
		while( left <= right)
		{
			mid = left + (right-left)/2 ;
			if(A[mid] == target)
				return mid;
			if( A[left] <= A[mid] ) //left side sorted, including mid, it is sorted in [l,mid]
			{
				if( A[mid] > target && A[left] <= target )
					right = mid - 1;
				else
					left = mid + 1;
			}
			else if ( A[left] > A[mid] ) //right side sorted, [mid,r] must be sorted, if [l,mid] not sorted
			{
				if ( A[mid] < target && A[right] >= target )
					left = mid + 1;
				else
					right = mid - 1;
			}
		}
		return -1;
    }
};

int main()
{
	int a , target;
	int A[N];
	for(int i=0; i<N; i++)
	{
		cin>>a;
		A[i] = a;
	}
	cin>>target;
	Solution lin;
	cout<<lin.search( A,N,target)<<endl;
}

//方法一：自测Accepted
//observation is the key, try to solve it by modifying binary search
class Solution {
public:
    int search(int A[], int n, int target) {
        if( n==0 )
			return -1;
        int mid = 0;
		int left=0, right=n-1;
		while( left <= right)
		{
			mid = left + (right-left)/2 ;
			if(A[mid] == target)
				return mid;
			if( A[left] <= A[mid] ) //left side sorted, including mid, it is sorted in [l,mid]
			{
				if( A[mid] > target && A[left] <= target )
					right = mid - 1;
				else
					left = mid + 1;
			}
			else if ( A[left] > A[mid] ) //right side sorted, [mid,r] must be sorted, if [l,mid] not sorted
			{
				if ( A[mid] < target && A[right] >= target )
					left = mid + 1;
				else
					right = mid - 1;
			}
		}
		return -1;
    }
};

//方法二：其他版本
class Solution {
public:
    int search(int A[], int n, int target) {
        int left = 0;
        int right = n-1;
        while(left < right)
        {
            int mid = left+(right-left)/2;
            if(A[mid] >= A[left])//left side sorted, including mid,
            {
                if(A[left] <= target && target <= A[mid]) right = mid;
                else left = mid+1;
            }
            else//right side sorted
            {
                if(A[mid] <= target && target <= A[right]) left = mid;
                else right = mid-1;
            }
        }
        if(right >= 0 && right < n && A[right] == target) return right;
        else return -1;
    }
};