算法33. Search in Rotated Sorted Array

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.

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

class Solution {
    public int search(int[] nums, int target) {
        
    }
}

说，假设有一个升序排序的数组，可能在某个点旋转。
找到目标值就返回下标，找不到返回-1。
可以假设没有重复的数字。
时间复杂度在O(log n)以内。

要在O(log n)以内首先想到进行折半查找，但是进行旋转后，就不是全部有序的了，这点要注意。

以下为代码：

public int search(int[] nums, int target) {
    if (nums == null || nums.length == 0) {
        return -1;
    }

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

            } else {
                right = mid - 1;
            }
        } else {
            if (nums[mid] >= nums[0]) {
                left = mid + 1;
            } else {
                if (target <= nums[n - 1]) {
                    left = mid + 1;
                } else {
                    right = mid - 1;
                }
            }
        }
    }

    return -1;
}