[Swift]LeetCode1067. 范围内的数字计数 | Digit Count in Range

★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★

➤微信公众号：山青咏芝（shanqingyongzhi）

➤博客园地址：山青咏芝（https://www.cnblogs.com/strengthen/ ）

➤GitHub地址：https://github.com/strengthen/LeetCode

➤原文地址：https://www.cnblogs.com/strengthen/p/10961687.html

➤如果链接不是山青咏芝的博客园地址，则可能是爬取作者的文章。

➤原文已修改更新！强烈建议点击原文地址阅读！支持作者！支持原创！

★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★

热烈欢迎，请直接点击！！！

进入博主App Store主页，下载使用各个作品！！！

注：博主将坚持每月上线一个新app！！！

Given an integer d between 0 and 9, and two positive integers low and high as lower and upper bounds, respectively. Return the number of times that d occurs as a digit in all integers between low and high, including the bounds low and high.

Example 1:

Input: d = 13
Output: 6
Explanation: 
The digit d=1 occurs 6 times in 1,10,11,12,13. Note that the digit d=1 occurs twice in the number 11.

Example 2:

Input: d = 250
Output: 35
Explanation: 
The digit d=3 occurs 35 times in 103,113,123,130,131,...,238,239,243.

Note:

1. 0 <= d <= 9
2. 1 <= low <= high <= 2×10^8

给定一个在 0 到 9 之间的整数 d，和两个正整数 low 和 high 分别作为上下界。返回 d 在 low 和 high 之间的整数中出现的次数，包括边界 low 和 high。

示例 1：

输入：d = 1, low = 1, high = 13
输出：6
解释： 
数字 d=1 在 1,10,11,12,13 中出现 6 次。注意 d=1 在数字 11 中出现两次。

示例 2：

输入：d = 3, low = 100, high = 250
输出：35
解释：
数字 d=3 在 103,113,123,130,131,...,238,239,243 出现 35 次。

提示：

1. 0 <= d <= 9
2. 1 <= low <= high <= 2×10^8

Runtime: 4 ms

Memory Usage: 20.7 MB

 1 class Solution {
 2     func digitsCount(_ d: Int, _ low: Int, _ high: Int) -> Int {
 3         let a:[Int] = f(high)
 4         let b:[Int] = f(low - 1)
 5         return a[d] - b[d]
 6     }
 7     
 8     func f(_ n:Int) -> [Int]
 9     {
10         var dev:[Int] = [Int](repeating:0,count:10)
11         if n == 0 {return dev}
12         var i:Int = 1
13         while(i <= n)
14         {
15             let a:Int = (n/i)/10
16             for j in 0..<10
17             {
18                 dev[j] += a*i
19             }
20             dev[0] -= i
21             for j in 0..<(n/i)%10
22             {
23                 dev[j] += i
24             }
25             dev[(n/i)%10] += (n%i) + 1
26             i *= 10
27         }
28         return dev
29     }
30 }