Lintcode: Recover Rotated Sorted Array

Given a rotated sorted array, recover it to sorted array in-place.

Example
[4, 5, 1, 2, 3] -> [1, 2, 3, 4, 5]

Challenge
In-place, O(1) extra space and O(n) time.

Clarification
What is rotated array:

    - For example, the orginal array is [1,2,3,4], The rotated array of it can be [1,2,3,4], [2,3,4,1], [3,4,1,2], [4,1,2,3]

我的做法是先扫描不是ascending order的两个点，然后reverse array三次，分别是(0, i), (i+1, num.length-1), (0, num.length-1)

 public class Solution {
     /**
      * @param nums: The rotated sorted array
      * @return: The recovered sorted array
      */
     public void recoverRotatedSortedArray(ArrayList<Integer> nums) {
         // write your code
         if (nums==null || nums.size()==0 || nums.size()==1) return;
         int i = 0;
         for (i=0; i<nums.size()-1; i++) {
             if (nums.get(i) > nums.get(i+1)) break;
         }
         if (i == nums.size()-1) return;
         reverse(nums, 0, i);
         reverse(nums, i+1, nums.size()-1);
         reverse(nums, 0, nums.size()-1);
     }

     public void reverse(ArrayList<Integer> nums, int l, int r) {
         while (l < r) {
             int temp = nums.get(l);
             nums.set(l, nums.get(r));
             nums.set(r, temp);
             l++;
             r--;
         }
     }
 }