[leetcode] Number of Digit One

From : https://leetcode.com/problems/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.

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Solution :

reference: https://leetcode.com/discuss/44281/4-lines-o-log-n-c-java-python
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 cnt = 0;
for (long long m = 1; m <= n; m *= 10) {
int a = n/m, b = n%m;
cnt += (a + 8) / 10 * m + (a % 10 == 1) * (b + 1);
}
return cnt;
}
};