33. 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.

使用二分查找。以4 5 6 7 0 1 2为例。若mid=5，则low=mid+1当且仅当targer在(nums[0] > target) ^ (nums[0] > nums[mid]) ^ (target > nums[mid])下成立

public class Solution {    public int search(int[] nums, int target) {        int lo = 0, hi = nums.length - 1;        while (lo < hi) {            int mid = (lo + hi) / 2;            if ((nums[0] > target) ^ (nums[0] > nums[mid]) ^ (target > nums[mid]))                lo = mid + 1;            else                hi = mid;        }        return lo == hi && nums[lo] == target ? lo : -1;    }}