33. Search in Rotated Sorted Array

解题思路：
1. 先找到最小值所在位置；
2. 以循环交界处（最小值前一位）为分界，分两段做二分查找

public class Solution {    public int binarySearch(int[] nums, int beg, int end, int target){        if (beg == end){            if (nums[beg] == target){                return beg;            }else{                return -1;            }        }else{            int mid = beg + (end - beg) / 2;            if (nums[mid] < target){                beg = mid + 1;            }else{                end = mid;            }            return binarySearch(nums, beg, end, target);        }    }    public int search(int[] nums, int target) {        int beg = 0, end = nums.length - 1, mid;        //先寻找最小值的位置        while (beg < end){            mid = beg + (end - beg) / 2;            if (nums[mid] > nums[end])                beg = mid + 1;            else                end = mid;        }        //nums_min = nums[beg],分两段二分查找        if (beg > 0){            int ret = binarySearch(nums, 0, beg - 1, target);            if (ret != -1)                return ret;        }        return binarySearch(nums, beg, nums.length - 1, target);    }}