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)

 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 }