[Swift]LeetCode902. 最大爲 N 的數字組合 | Numbers At Most N Given Digit Set

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We have a sorted set of digits D, a non-empty subset of {'1','2','3','4','5','6','7','8','9'}.  (Note that '0' is not included.)

Now, we write numbers using these digits, using each digit as many times as we want.  For example, if D = {'1','3','5'}, we may write numbers such as '13', '551', '1351315'.

Return the number of positive integers that can be written (using the digits of D) that are less than or equal to N. 

Example 1:

Input: D = ["1","3","5","7"], N = 100 Output: 20 Explanation: The 20 numbers that can be written are: 1, 3, 5, 7, 11, 13, 15, 17, 31, 33, 35, 37, 51, 53, 55, 57, 71, 73, 75, 77. 

Example 2:

Input: D = ["1","4","9"], N = 1000000000 Output: 29523 Explanation: We can write 3 one digit numbers, 9 two digit numbers, 27 three digit numbers, 81 four digit numbers, 243 five digit numbers, 729 six digit numbers, 2187 seven digit numbers, 6561 eight digit numbers, and 19683 nine digit numbers. In total, this is 29523 integers that can be written using the digits of D. 

Note:

1. D is a subset of digits '1'-'9' in sorted order.
2. 1 <= N <= 10^9

---

咱們有一組排序的數字 D，它是  {'1','2','3','4','5','6','7','8','9'} 的非空子集。（請注意，'0' 不包括在內。）

如今，咱們用這些數字進行組合寫數字，想用多少次就用多少次。例如 D = {'1','3','5'}，咱們能夠寫出像 '13', '551', '1351315' 這樣的數字。

返回能夠用 D 中的數字寫出的小於或等於 N 的正整數的數目。 

示例 1：

輸入：D = ["1","3","5","7"], N = 100
輸出：20
解釋：
可寫出的 20 個數字是：
1, 3, 5, 7, 11, 13, 15, 17, 31, 33, 35, 37, 51, 53, 55, 57, 71, 73, 75, 77.

示例 2：

輸入：D = ["1","4","9"], N = 1000000000
輸出：29523
解釋：
咱們能夠寫 3 個一位數字，9 個兩位數字，27 個三位數字，
81 個四位數字，243 個五位數字，729 個六位數字，
2187 個七位數字，6561 個八位數字和 19683 個九位數字。
總共，可使用D中的數字寫出 29523 個整數。 

提示：

1. D 是按排序順序的數字 '1'-'9' 的子集。
2. 1 <= N <= 10^9

---

Runtime: 8 ms

Memory Usage: 19.6 MB

 1 class Solution {
 2     func atMostNGivenDigitSet(_ D: [String], _ N: Int) -> Int {
 3         var NS:String = String(N)
 4         var digit:Int = NS.count
 5         var dsize:Int = D.count
 6         var rtn:Int = 0
 7         for i in 1..<digit
 8         {
 9             rtn += Int(pow(Double(dsize), Double(i)))
10         }
11         for i in 0..<digit
12         {
13             var hasSameNum:Bool = false
14             for d in D
15             {
16                 if d[0] < NS[i]
17                 {
18                     rtn += Int(pow(Double(dsize), Double(digit - i - 1)))
19                 }
20                 else if d[0] == NS[i]
21                 {
22                     hasSameNum = true
23                 }
24             }
25             if !hasSameNum {return rtn}
26         }
27         return rtn + 1
28     }
29 }
30 
31 //String擴展
32 extension String {        
33     //subscript函數能夠檢索數組中的值
34     //直接按照索引方式截取指定索引的字符
35     subscript (_ i: Int) -> Character {
36         //讀取字符
37         get {return self[index(startIndex, offsetBy: i)]}
38     }
39 }

---

Runtime: 8 ms

Memory Usage: 20.2 MB

 1 class Solution {
 2     func atMostNGivenDigitSet(_ D: [String], _ N: Int) -> Int {
 3         let nums = D.map { (item) -> Int in return Int(item)! }.sorted()
 4         var number = N / 10
 5         var digitNumber = 1
 6         var disgits = [N % 10]
 7         while number != 0 {
 8             digitNumber += 1
 9             disgits.insert(number % 10, at: 0)
10             number = number / 10
11             
12         }
13         var count = 0
14         for i in 1..<digitNumber {
15             count += Int(pow(Double(nums.count), Double(i)))
16         }
17         count += hightDigitNumber(nums: nums, disgits: disgits)
18         for i in 0..<disgits.count {
19             var isBreak = true
20             for num in nums {
21                 if num == disgits[i] {
22                     if i == disgits.count - 1 {
23                         count += 1
24                     }else {
25                         count += hightDigitNumber(nums: nums, disgits: Array(disgits[i + 1..<disgits.count]))
26                     }
27                     isBreak = false
28                 }
29             }
30             if isBreak {
31                 break;
32             }
33         }
34         return count
35     }
36     func hightDigitNumber(nums:[Int],disgits:[Int]) -> Int {
37         var lessCount = 0
38         for num in nums {
39             if num < disgits[0] {
40                 lessCount += 1
41             }
42         }
43         return Int(pow(Double(nums.count), Double(disgits.count - 1))) * lessCount
44     }
45 }