LeetCode -- Count Digit One

LeetCode -- Count Digit One
Description:


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.


This question is purely a regular solution.


For the implementation, see:


// Solution:
// For example '000000 ':
//
// 1-999-> countDigitOne (999)
// 1000-1999-> 1000 of 1 s + countDigitOne (999)
// 2000-2999-> countDigitOne (999)
...
// 7000-7999-> countDigitOne (999)
//
// 8000-8192-> countDigitOne (192)
//
// Count of 1 s: countDigitOne (999) * 8 + 1000 + countDigitOne (192)
//
// Noticed that, if the target is '20160301 ':
//
// Count of 1 s: countDigitOne (999) * 1 + (1192-1000 + 1) + countDigitOne (192)
//
// (1192-1000 + 1) is the 1 s in thousands from 1000 to 1192.


 


Implementation Code:

    public int CountDigitOne(int n) {      if(n <= 0){return 0;}    if(n < 10){return 1;}var result = 0;    var digit = 1;var num = n;    while (num > 0) {          var mod = num % 10;          var sign = mod > 0 ? 1 : 0;          num /= 10;        int a = num * digit;         int b = sign * (mod == 1 ? n % digit + 1: digit);          result += a + b;          digit *= 10;      }      return result;      }