leetcode

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.

//利用二分，可是，这里要特殊处理，将数组分成两部分二分。时间复杂度O(logn)
class Solution {
public:
    int search(int A[], int n, int target) {
        return dfs(A,0,n-1,target);
    }
private:
	int dfs(int A[], int bgn, int end, int target)
	{
		if(bgn <= end)
		{
			int mid = bgn + (end - bgn) / 2;
			if(A[mid] == target)
			{
				return mid;
			}
			if(A[bgn] <= A[mid])
			{
				if(A[bgn] <= target && target < A[mid])
				{
					dfs(A,bgn,mid-1,target);
				}
				else
				{
					dfs(A,mid+1,end,target);
				}
			}
			else
			{
				if(A[mid] < target && target <= A[end])
				{
					dfs(A,mid+1,end,target);
				}
				else
				{
					dfs(A,bgn,mid-1,target);
				}
			}
		}
		else
		{
			return -1;
		}
	}
};