3. Digit Counts【medium】

Count the number of k's between 0 and n. k can be 0 - 9.

 
Example

if n = 12, k = 1 in

[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12]

we have FIVE 1's (1, 10, 11, 12)

解法一:

 1 class Solution {
 2  public:
 3     // param k : description of k
 4     // param n : description of n
 5     // return ans a integer
 6     int digitCounts(int k, int n) {
 7         int count = 0;
 8         if (k == 0) {
 9             count = 1;
10         }
11         for (int i = 1; i <= n; i++) {
12             int number = i;
13             while (number > 0) {
14                 if (number % 10 == k) {
15                     count += 1;
16                 } 
17                 number /= 10;
18             }
19         }
20         
21         return count;
22     }
23 };
原文地址:https://www.cnblogs.com/abc-begin/p/8157964.html