233. Number of Digit One

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.

Hint:

Beware of overflow.

//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 的总个数（个位,十位, 百位）

//以算百位上1为例子:   假设百位上是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:m表示位数
//(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;
long b = n % m;
ones += (a + 8) / 10 * m + (a % 10 == 1) * (b + 1);
}
return ones;
}
};