LeetCode -- Count Digit One

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。

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.

链接：
https://leetcode.com/discuss/44496/5lins-solution-using-recursion

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;
    }

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