Leetcode: Lexicographical Numbers

Given an integer n, return 1 - n in lexicographical order.

For example, given 13, return: [1,10,11,12,13,2,3,4,5,6,7,8,9].

Please optimize your algorithm to use less time and space. The input size may be as large as 5,000,000.

Solution 1:

If we look at the order we can find out we just keep adding digit from 0 to 9 to every digit and make it a tree.
Then we visit every node in pre-order.
       1        2        3    ...
      /\        /\       /\
   10 ...19  20...29  30...39   ....

 public class Solution {
     public List<Integer> lexicalOrder(int n) {
         ArrayList<Integer> res = new ArrayList<Integer>();
         for (int i=1; i<=9; i++) {
             helper(res, i, n);
         }
         return res;
     }

     public void helper(ArrayList<Integer> res, int cur, int n) {
         if (cur > n) return;
         res.add(cur);
         for (int i=0; i<=9; i++) {
             helper(res, cur*10+i, n);
         }
     }
 }

Solution 2:

O(N) time, O(1) space

The basic idea is to find the next number to add.
Take 45 for example: if the current number is 45, the next one will be 450 (450 == 45 * 10)(if 450 <= n), or 46 (46 == 45 + 1) (if 46 <= n) or 5 (5 == 45 / 10 + 1)(5 is less than 45 so it is for sure less than n).
We should also consider n = 600, and the current number = 499, the next number is 5 because there are all "9"s after "4" in "499" so we should divide 499 by 10 until the last digit is not "9".

Note: 第二、三种情况不能合并的原因是：不一定是因为最后一位是9才需要/10，有可能是因为curr+1>n

 public List<Integer> lexicalOrder(int n) {
         List<Integer> list = new ArrayList<>(n);
         int curr = 1;
         for (int i = 1; i <= n; i++) {
             list.add(curr);
             if (curr * 10 <= n) {
                 curr *= 10;
             } else if (curr % 10 != 9 && curr + 1 <= n) {
                 curr++;
             } else {
                 while ((curr / 10) % 10 == 9) {
                     curr /= 10;
                 }
                 curr = curr / 10 + 1;
             }
         }
         return list;
     }