LeetCode -- Count Digit One

LeetCode -- Count Digit One

題目描述：


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.


本題目純粹是找規律求解。


該實現參考了：


//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.


 


實現代碼：

    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;  
    }