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.

Answer:

Lets see some pattern.

For n <= 0 return 0;

For n from 1 - 9 return 1;

For n from 1 - 99 divide by two cases,

A.the last digit is 1, so there are 1 11 21 31 41 51 61 71 81 91, 10 numbers in total.

B. the second digit is 1, there are 11, 12....19.

so in total 20 numbers.

We just need to compute the 1 in each digit and that is fine.

for example n = 35601.

for the last digit, there are from 00001, 00011,00021,..........35601 , so 3561 in total.

for the second last digit, there are 0000X, 0001X, ....3550X    and 3560Y

for x could be from 0-9, there are 356 * 10 ones from 0000X to 3550X,

for y could be 0 and 1, there are 2 ones for 3560y.

for the third last digit, since 6 is bigger than 1, there are 001XX ---->351XX  in total 36 * 100 numbers.

for the fourth last digit, there are 01XXX ---> 31XXX in total 4 * 1000 ones.

for the first digit, we can view the number as 035601, and it is from 01XXXX -> 01XXXX in total 10000 ones.

Adding 3561 + 2560 + 2 + 3600 + 4000 + 10000 is the answer.

abcdefg,

if we want see how many ones in the 'e' digit,

if(e > 1) so the ones comes from 00001XX -> abcd1XX it is (abcd  + 1) * 100 numbers, since there would be 0 -> 99 , 100 numbers represented by XX

if(e == 0), the ones comes from 00001XX -> (abcd - 1)1XX, the ones are abcd * 100.

if(e == 1) the ones comes from 00001XX -> (abcd - 1)1XX  and abcd1XX where abcd1XX <= abcd1fg, so XX represents fg + 1 numbers, which is in total (abcd * 100)+ (fg + 1)

Code:

public class Solution {    public int countDigitOne(int n) {        if(n <= 0) return 0;        if(n < 10) return 1;        int ret = 0;                int factor = 1;        int num = n;        while(num > 0){            int digit = num % 10;            int pre = num / 10;            ret += factor * pre;            if(digit > 1){                ret += factor;            } else if(digit == 1){                ret = ret + n % factor + 1;            }            num = pre;            factor *= 10;        }        return ret;    }}