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.

方法1：

利用排序，把数字按照lexicographical order 排序。但是最后通不过。

 public class Solution {
     public List<Integer> lexicalOrder(int n) {
         List<Integer> list = new ArrayList<>();
         if (n < ) return list;
         for (int i = ; i <= n; i++) {
             list.add(i);
         }

         Collections.sort(list, new Comparator<Integer>() {
             @Override
             public int compare(Integer i1, Integer i2) {
                 return String.valueOf(i1).compareTo(String.valueOf(i2));
             }
         });
         return list;
     }
 }

方法2：来自leetcode discuss  https://discuss.leetcode.com/topic/55377/simple-java-dfs-solution

The idea is pretty simple. 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) {
         List<Integer> res = new ArrayList<>();
         for (int i = ; i < ; ++i) {
             dfs(i, n, res);
         }
         return res;
     }

     public void dfs(int cur, int n, List<Integer> res) {
         if (cur > n) return;
         res.add(cur);
         for (int i = ; i < ; ++i) {
             dfs( * cur + i, n, res);
         }
     }
 }