Search in Rotated Sorted Array

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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.

int binaryfind(int* A, int first,int last, int target){
	while (first<=last){
		int mid = (first + last) / 2;
		if (A[mid] == target)
			return mid;
		else if (A[mid] < target)
			first = mid + 1;
		else
			last = mid - 1;
	}
	return -1;
}
int search(int A[], int n, int target) {
	int first = 0;
	int last = n - 1;
	if (A[first] <= A[last])
		return binaryfind(A, first, last, target);
	int idx = -1;
	while (first<=last){
		int mid = (first + last) / 2;
		if (A[first] <= A[mid]){
			if (target <= A[mid]){                    //注意等号的限定条件
				idx = binaryfind(A, first, mid, target);
				if (idx!=-1)
					break;
			}	
			first = mid + 1;			
		}
		else{
			if (target >= A[mid]){
				idx = binaryfind(A, mid, last, target);
				if (idx != -1)
					break;
			}	
			last = mid - 1;
		}
	}
	return idx;	
}

Search in Rotated Sorted Array

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原文地址：http://blog.csdn.net/li_chihang/article/details/44403195