LeetCode Search in Rotated Sorted Array

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.
有序数组采用二分搜索进行查找。
如果start--middle是递增的

       如果target在其中，在start--middle中搜索

       否则在middle--end中搜索

如果middle--end是递增的

       如果target在其中，在middle--end中搜索

       否则在start--middle中搜索

public class Solution {
public int search(int[] A, int target) {
int middle = 0;
int start = 0;
int end = A.length;
while (true) {
middle = (start + end) / 2;
if (start >= end)
return -1;
if (A[middle] == target)
return middle;
if(A[start]<A[middle])
{
if(A[start]<=target&&A[middle]>target)
end=middle;
else start=middle+1;
}
else {
if(A[middle]<target&&A[end-1]>=target)
start=middle+1;
else end=middle;
}
}
}
}