Search in rotated sorted array

Solution: 

In a sorted rotated array, half of the elements must be in a sorted sequence. 

So we can compare the element at begin, medium, end to figure out which part is in a sorted sequence. Then we can tell whether the target element is in the sorted half, if not , other half.

Since each time we discard half of the element. the time complexity is O(logN)

time: O(logN), space : O(1)

题意是让你从一个旋转过后的递增序列中寻找给定值，找到返回索引，找不到返回-1，我们在下面这组数据中寻找规律。

1 2 4 5 6 7 0

2 4 5 6 7 0 1

4 5 6 7 0 1 2

5 6 7 0 1 2 4

6 7 0 1 2 4 5

7 0 1 2 4 5 6

由于是旋转一次，所以肯定有一半及以上的序列仍然是具有递增性质的，我们观察到如果中间的数比左面的数大的话，那么左半部分序列是递增的，否则右半部分就是递增的，那么我们就可以确定给定值是否在递增序列中，从而决定取舍哪半边。

https://github.com/Blankj/awesome-java-leetcode/blob/master/note/033/README.md