[LintCode] #39 Recover Rotated Sorted Array

1) in place == re-sort O(nlogn)
2) O(n) space, O(n) time
3) 3 time reverse, O(1) space, O(n) time

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