@H_301_1@Suppose a sorted array is rotated at some pivot unknown to you beforehand.@H_404_3@
(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.@H_404_3@
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
原文链接：https://www.f2er.com/javaschema/284155.html