Search in Rotated Sorted Array

本文介绍了一种在已知没有重复元素的旋转数组中查找特定目标值的方法。文章详细阐述了如何通过二分查找来确定旋转点,并在此基础上实现有效的搜索策略。

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;	
}


 

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