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:

1. Beware of overflow.

public class Solution {    public int countDigitOne(int n) {        int cnt = 0;                for(long k=1; k<=n; k*=10) {        long r = n / k, m = n % k;                // r / 10是该位左边的影响，* k 是该位右边的影响        // cnt += r / 10 * k;                // 该位正好为1时就只算散的        // cnt += r / 10 * k + (r % 10 == 1 ? m+1 : 0);                // 该位大于1，要加8进一位        cnt += (r+8) / 10 * k + (r % 10 == 1 ? m+1 : 0);        }                return cnt;    }}

主要参考https://discuss.leetcode.com/topic/18972/ac-short-java-solution

Solution explanation:

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 etc), we know that there's a one at ones' 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.

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

The answer is (n / k) / 10.

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 have a formula to count the ones at k's place: r / 10 * k

2). So far, everything looks good, but we need to 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 ones's place, perfect, but for tens' place, we get (n / 10) / 10 * 10 = 0, that's not right, there should be a one at tens' 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). Wait, how about 20, 21 and 22?

Let's say 20, use the 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 10 more ones to the tens' place, a clever way to fix is to add 8 to r, so our final formula is:

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

As you can see, it's all about how we fix the formula.