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 < 1) return list;
        for (int i = 1; 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 = 1; i < 10; ++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 = 0; i < 10; ++i) {
            dfs(10 * cur + i, n, res);
        }
    }
}