Given an integer n, count the total number of digit 1 appearing in all non-negative integers less than or equal to n.

 

Example:

Input: 13
Output: 6 
Explanation: Digit 1 occurred in the following numbers: 1, 10, 11, 12, 13.

 

Approach #1: Math. [Java]

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

　　

Analysis:

Let's start by counting the ones for every 10 numbers....

0, 1, 2, 3, ...., 9 (1)

10, 11, 12, 13, ...., 19 (1) + 10

20, 21, 22, 23, ...., 29 (1)

...

90, 91, 92, 93, ...., 99 (1)

100, 101, 102, 103, ...., 109 (10 + 1)

110, 111, 112, 113, ...., 119 (10 + 1) + 10

120, 121, 122, 123, ...., 129 (10 + 1)

...

190, 191, 192, 193, ...., 199 (10 + 1)

1. If we don't look at those special rows (start with 10, 110 ect), we know that there's a one at one's place in every 10 numbers, there are 10 ones at tens' place in every 100 numbers, and 100 ones at hundreds' place in every 1000 numbers, so on and so forth.

Let's start with ones' place and count how many ones at this place, set k = 1, as mentioned above, there is a one at one's place in every 10 numbers, so how many 10 numbers do we have?

The answer is (n / k) / 10 * k.

Now let's count the ones in tens' place, set k = 10, as mentioned above, there are 10 ones at tens' place in every 100 numbers, so how many 100 numbers do we have?

The answer is (n / k) / 10, and the number of ones at ten's place is (n / k) / 10 * k.

Let r = n / k, now we a formula to count the ones at k's place: r / 10 * k.

 

2. So far, everything looks good, but we need o fix those special rows, how?

We can use the mod. Take 10, 11, and 12 for example, if n is 10, we get (n / 1) / 10 * 1 = 1 ones at one's place, perfect, but for tens' place, we get (n / 10) / 10 * 10 = 0, that's not right, there should be a one at ten's place! Calm down, from 10 to 19, we always have a one at tens's place, let m = n % k, the number of ones at this special place is m + 1, so let's fix the formula to be:

r / 10 * k + (r % 10 == 1 ? m + 1 : 0)

 

3. How about 20, 21, and 22?

Let's say 20, use above formula we get 0 ones at tens' place, but it should be 10! How to fix it? We know that once the digit is larger than 2, we should add 8 to r, so our formula is :

(r + 8) / 10 * k + (r % 10 == 1 ? m + 1 : 0)

 

Reference:

https://leetcode.com/problems/number-of-digit-one/discuss/64390/AC-short-Java-solution