[Lintcode]Recover Rotated Sorted Array Show result

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.

找到数组的旋转点，旋转点左侧倒序排列，右侧同样倒序排列，最后整个数组再倒序排列。

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




public class Solution {    /**     * @param nums: The rotated sorted array     * @return: void     */    public void recoverRotatedSortedArray(ArrayList<Integer> nums) {                for(int i = 0; i < nums.size() - 1; i++) {            if(nums.get(i) > nums.get(i + 1)) {                reverse(nums, 0, i);                reverse(nums, i + 1, nums.size() - 1);                reverse(nums, 0, nums.size() - 1);                return;            }        }    }        private void reverse(ArrayList<Integer> nums, int start, int end) {        while(start < end) {            int tmp = nums.get(start);            nums.set(start, nums.get(end));            nums.set(end, tmp);            start ++;            end --;        }    }}