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。

 1 public class Solution {
 2     public List<Integer> lexicalOrder(int n) {
 3         List<Integer> list = new ArrayList<>();
 4         if (n < 1) return list;
 5         for (int i = 1; i <= n; i++) {
 6             list.add(i);
 7         }
 8 
 9         Collections.sort(list, new Comparator<Integer>() {
10             @Override
11             public int compare(Integer i1, Integer i2) {
12                 return String.valueOf(i1).compareTo(String.valueOf(i2));
13             }
14         });
15         return list;
16     }
17 }

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   ....
 1 public class Solution {
 2     public List<Integer> lexicalOrder(int n) {
 3         List<Integer> res = new ArrayList<>();
 4         for (int i = 1; i < 10; ++i) {
 5             dfs(i, n, res);
 6         }
 7         return res;
 8     }
 9 
10     public void dfs(int cur, int n, List<Integer> res) {
11         if (cur > n) return;
12         res.add(cur);
13         for (int i = 0; i < 10; ++i) {
14             dfs(10 * cur + i, n, res);
15         }
16     }
17 }

 

posted @ 2017-01-09 11:40  北叶青藤  阅读(185)  评论(0编辑  收藏  举报