Search in Rotated Sorted Array 倒置数组中的二分查找

题目：

Suppose a sorted array is rotated at some pivot unkNown to you beforehand.

(i.e., 0 1 2 4 5 6 7might become4 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.

解答代码：

int binarySearch(int A[],int start,int end,int key)
{
	int mid;
	while (start <= end)
	{
		mid = (start + end) / 2;
		if (A[mid] == key)
		{
			return mid;
		}
		if (A[start] <= A[mid])
		{
			if (A[start] <= key && key < A[mid])
				end = mid;
			else
				start = mid + 1;
		}
		else
		{
			if (A[mid] < key && key <= A[end])
				start = mid + 1;
			else
				end = mid;
		}
	}
	return -1;
}

解答要点：找到mid后分为两半处理，其中必然有一半是有序的，判断key是否在这有序数组中，如果在则继续在这数组中查找，否则在另一数组中查找

原文地址：https://www.jb51.cc/javaschema/284155.html