题目描述：
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.
本题目纯粹是找规律求解，
LeetCode Count Digit One
。该实现参考了：
//Solution:
//For example '8192':
//
//1-999 -> countDigitOne(999)
//1000-1999 -> 1000 of 1s + countDigitOne(999)
//2000-2999 -> countDigitOne(999)
...
//7000-7999 -> countDigitOne(999)
//
//8000-8192 -> countDigitOne(192)
//
//Count of 1s : countDigitOne(999)*8 + 1000 + countDigitOne(192)
//
//Noticed that, if the target is '1192':
//
//Count of 1s : countDigitOne(999)*1 + (1192 - 1000 + 1) + countDigitOne(192)
//
//(1192 - 1000 + 1) is the 1s in thousands from 1000 to 1192.
实现代码：
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; }