leetcode 233 Number of Digit One

问题：从1到n的整数中1出现的次数

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.

代码

/*
intuitive: 每10个数, 有一个个位是1, 每100个数, 有10个十位是1, 每1000个数, 有100个百位是1.  做一个循环, 每次计算单个位上1得总个数(个位,十位, 百位).  
例子:
以算百位上1为例子:   假设百位上(m=100)是0, 1, 和 >=2 三种情况: 
    case 1: n=3141092, a= 31410, b=92. 计算百位上1的个数应该为 3141 *100 次.
    case 2: n=3141192, a= 31411, b=92. 计算百位上1的个数应该为 3141 *100 + (92+1) 次. 
    case 3: n=3141592, a= 31415, b=92. 计算百位上1的个数应该为 (3141+1) *100 次. 
以上三种情况可以用 一个公式概括:
(a + 8) / 10 * m + (a % 10 == 1) * (b + 1);
*/

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