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]ChallengeIn-place, O(1) extra space and O(n) time.ClarificationWhat 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)

 1 public class Solution { 2     /** 3      * @param nums: The rotated sorted array 4      * @return: The recovered sorted array 5      */ 6     public void recoverRotatedSortedArray(ArrayList<Integer> nums) { 7         // write your code 8         if (nums==null || nums.size()==0 || nums.size()==1) return; 9         int i = 0;10         for (i=0; i<nums.size()-1; i++) {11             if (nums.get(i) > nums.get(i+1)) break;12         }13         if (i == nums.size()-1) return;14         reverse(nums, 0, i);15         reverse(nums, i+1, nums.size()-1);16         reverse(nums, 0, nums.size()-1);17     }18     19     public void reverse(ArrayList<Integer> nums, int l, int r) {20         while (l < r) {21             int temp = nums.get(l);22             nums.set(l, nums.get(r));23             nums.set(r, temp);24             l++;25             r--;26         }27     }28 }