Leetcode: Search in Rotated Sorted Array

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本文介绍了一种在旋转排序数组中查找目标值的算法实现。该算法通过调整二分搜索法来适应旋转数组的特点，能够在O(log n)的时间复杂度内找到目标元素的位置，若未找到则返回-1。文章中的代码示例清晰地展示了这一过程。

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 search(int A[], int n, int target) {
        // Note: The Solution object is instantiated only once.
        int begin = 0;
		int end = n-1;
		while(begin < end)
		{
			int mid = (begin+end)/2;
			if(A[mid] == target)return mid;
			else if(A[begin] <= A[mid])
			{
				if(A[begin] <= target && target < A[mid])
					end = mid-1;
				else
					begin = mid+1;
			}
			else
			{
				if(A[mid] < target && target <= A[end])
					begin = mid+1;
				else
					end = mid-1;
			}
		}
		if(begin==end && A[begin]==target)
			return begin;
		else
			return -1;
    }