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=0;
        for(long long m=1;m<=n;m*=10)
        {
            int a=n/m,b=n%m;
            if(a%10==0)
                cnt+=a/10*m;
            else if(a%10==1)
                cnt+=a/10*m+(b+1);
            else
                cnt+=(a/10+1)*m;
        }
        return cnt;
    }
};

方法二

class Solution {
public:
    int countDigitOne(int n) {
        int cnt=0;
        for(long long m=1;m<=n;m*=10)
            cnt=cnt+(n/m+8)/10*m+(n/m%10==1)*(n%m+1);
        return cnt;
    }
};