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 ....
1 public class Solution { 2 public List<Integer> lexicalOrder(int n) { 3 ArrayList<Integer> res = new ArrayList<Integer>(); 4 for (int i=1; i<=9; i++) { 5 helper(res, i, n); 6 } 7 return res; 8 } 9 10 public void helper(ArrayList<Integer> res, int cur, int n) { 11 if (cur > n) return; 12 res.add(cur); 13 for (int i=0; i<=9; i++) { 14 helper(res, cur*10+i, n); 15 } 16 } 17 }
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
1 public List<Integer> lexicalOrder(int n) { 2 List<Integer> list = new ArrayList<>(n); 3 int curr = 1; 4 for (int i = 1; i <= n; i++) { 5 list.add(curr); 6 if (curr * 10 <= n) { 7 curr *= 10; 8 } else if (curr % 10 != 9 && curr + 1 <= n) { 9 curr++; 10 } else { 11 while ((curr / 10) % 10 == 9) { 12 curr /= 10; 13 } 14 curr = curr / 10 + 1; 15 } 16 } 17 return list; 18 }