[Swift]LeetCode386. 字典序排數 | Lexicographical Numbers

原文地址：http://www.javashuo.com/article/p-exwabrmn-q.html 

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.

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給定一個整數 n, 返回從 1 到 n 的字典順序。

例如，

給定 n =1 3，返回 [1,10,11,12,13,2,3,4,5,6,7,8,9] 。

請儘量的優化算法的時間複雜度和空間複雜度。 輸入的數據 n 小於等於 5,000,000。

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112ms

 1 class Solution {
 2     func lexicalOrder(_ n: Int) -> [Int] {
 3         var res:[Int] = [Int]()
 4         for i in 1...9
 5         {
 6             helper(i, n, &res)
 7         }
 8         return res
 9     }
10     
11     func helper(_ cur:Int,_ n:Int,_ res:inout[Int])
12     {
13         if cur > n {return}
14         res.append(cur)
15         for i in 0...9
16         {
17             if cur * 10 + i <= n
18             {
19                 helper(cur * 10 + i, n, &res)
20             }
21             else
22             {
23                 break
24             }
25         }
26     }
27 }

---

152ms

 1 class Solution {
 2     func lexicalOrder(_ n: Int) -> [Int] {
 3         var res = [Int]()
 4         for i in 1 ... 9 {
 5             dfs(i, n, &res)
 6         }
 7         
 8         return res
 9     }
10     
11     func dfs(_ cur: Int, _ n: Int, _ res: inout [Int]) {
12         if cur > n {
13             return
14         }
15         res.append(cur)
16         for i in 0 ... 9 {
17             if cur * 10 + i > n {
18                 return
19             }
20             dfs(cur * 10 + i, n, &res)
21         }
22     }
23 }

---

184ms

 1 class Solution {
 2     func lexicalOrder(_ n: Int) -> [Int] {
 3         var result = [Int]()
 4         var current = 1
 5         for i in 1 ... n {
 6             result.append(current)
 7             if current * 10 <= n {
 8                 current = current * 10
 9             } else if current % 10 != 9 && current + 1 <= n {
10                 current += 1
11             } else {
12                 while (current / 10) % 10 == 9 {
13                     current = current / 10
14                 }
15                 current = current / 10 + 1
16             }
17         }
18         return result
19     }
20 }