LeetCode 233 Number of Digit One 某一范围内的整数包含1的数量

Given an integer n, count the total number of digit 1 appearing in all non-negative integers less than or equal to n.
For example:
Given n = 13,
Return 6, because digit 1 occurred in the following numbers: 1, 10, 11, 12, 13.

方法一

 class Solution {
 public:
     int countDigitOne(int n) {
         int cnt=;
         for(long long m=;m<=n;m*=)
         {
             int a=n/m,b=n%m;
             if(a%==)
                 cnt+=a/*m;
             else if(a%==)
                 cnt+=a/*m+(b+);
             else
                 cnt+=(a/+)*m;
         }
         return cnt;
     }
 };

方法二

 class Solution {
 public:
     int countDigitOne(int n) {
         int cnt=;
         for(long long m=;m<=n;m*=)
             cnt=cnt+(n/m+)/*m+(n/m%==)*(n%m+);
         return cnt;
     }
 };