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

难度：medium

题目：假设一个按升序排序的数组在某个未知的轴上旋转。

给定一个搜索值，如果能找到则返回其所在数组中的位置，否则返回－1. 假定数组中无重复元素。算法时间复杂度要为O(log n)

思路：二叉搜索

Runtime: 10 ms, faster than 21.83% of Java online submissions for Search in Rotated Sorted Array.

Memory Usage: 26.8 MB, less than 41.44% of Java online submissions for Search in Rotated Sorted Array.

class Solution {    public int search(int[] nums, int target) {        int left = 0;         int right = nums.length - 1;        while (left <= right) {            int mid = left + (right - left) / 2;            if (target == nums[mid]) {                return mid;            }                        // left < mid  right            if (nums[left] < nums[mid]) {                if (target > nums[mid] || target < nums[left]) {                    left = mid + 1;                } else {                    right = mid - 1;                }            } else if (nums[left] > nums[mid]) { // left > mid  right                if (target < nums[mid] || target > nums[right]) {                    right = mid - 1;                } else {                    left = mid + 1;                }            } else { // left = mid                left += 1;            }        }                return -1;    }}