33. Search in Rotated Sorted Array 二分法

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.

这道题本来用for循环来做。。发现速度只打败了27%太慢了，所以改用二分，这道题可以说是二分的变形，因为从

public class Solution {    public int search(int[] nums, int target) {        int begin = 0;        int end = nums.length -1 ;        int mid = 0;        while(begin <= end){            mid = begin+(end-begin) / 2;            if(nums[mid] == target){                return mid;            }else if(nums[begin] > nums[mid]){                 // 这里不可以是 >=                 if(target > nums[mid] && target <= nums[end]){  // 这里不可以是 >= ,要<=                    begin = mid +1;                }else{                    end = mid - 1;                }            }else if(nums[begin] <= nums[mid]){               //  这里是 <=                if(target >= nums[begin] && target < nums[mid]){                    end = mid -1;                }else{                    begin = mid + 1;                }            }        }        return -1;         }}