Search in Rotated Sorted Array

Suppose an array sorted in ascending order 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.

Your algorithm's runtime complexity must be in the order of O(log n).

Example 1:

Input: nums = [4,5,6,7,0,1,2], target = 0
Output: 4

Example 2:

Input: nums = [4,5,6,7,0,1,2], target = 3
Output: -1

public int search(int[] nums, int target) {
        boolean isBig = false;
		
		if (nums.length == 0) {
			return -1;
		}
		int boundary = nums[0];

		if(target > boundary) {
			isBig = true;
		} else if(target == boundary) {
			return 0;
		}
		
		int begin = 0, end = nums.length - 1;
		int temp = 0;

		for (; end >= begin;) {
			temp = (begin + end) >> 1;
			if (target > nums[temp]) {
				if(isBig && boundary > nums[temp]) {
					end = temp - 1;
				} else {
					begin = temp + 1;
				}
			} else if (target < nums[temp]) {
				if(!isBig && nums[temp] >= boundary) {
					begin = temp + 1;
				} else {
					end = temp - 1;
				}	
			} else {
				return temp;
			}
		}
		return -1;
    }

我的代码逻辑：

以第一个数4作为边界点数：

假设一个数组：

 [4,5,6,7,0,1,2], 比4大的4-7叫做大串， 比4小的0-2叫做小串

由于效率原因，我们使用二分查找，其实就是不断缩小begin 和 end的过程，

1.当一个数大于4那就肯定在大串当中，注意一点， 什么时候开始缩小end呢，就是比如你要找7， 但是二分的值是0时，这个数比边界点数要小，说明就是缩小end了。

2.当一个数小于4那就肯定在小串当中，同理， 什么时候开始增加begin呢，就是比如你要找1， 但是二分的值是7时，这个数比边界点数要大，说明就是增加begin了。