Count 1 between 0 and N

Problem Description

Given a decimal integer n, the number of "1" occurrences in all integers from 1 to n is calculated.  

For example: 1 "1" appeared in n=2.

n=12 when 1,2,3,4,5,6,7,8,9,10,11,12. There were 5 "1".

Method one brute force solution

The most direct way is to start from 1 to N, each of which contains the number of "1" added together, the problem is solved.

The code is as follows:

publicclassCountOnes {

publiclong CountOnes(longn){

longi = 0,j = 1;

longcount = 0;

for(i = 1; i <= n; i++)

{

j = i;

while(j != 0)

{

if(j % 10== 1)

count++;

j = j / 10;

}

}

returncount;

}

publicstatic voidmain(String args[]){

CountOnes c = newCountOnes();

System.out.println(c.CountOnes(12));

System.out.println(c.CountOnes(120));

}

}

The time complexity of the algorithm is O (N*LGN)

Solution Two

1-digit situation:

In the solution two has been analyzed, greater than or equal to 1, there are 1, less than 1 is not.

2-digit situation:

N=13, the number of single-digit occurrences of 1 is 2, respectively 1 and 11, 10 digits 1, respectively, is 10,11,12,13, so f (N) = 2+4.

N=23, the number of single-digit occurrences of 1 is 3, respectively, 1,11,21, 10 digits 1 of the number of 10, respectively, 10~19,f (N) =3+10.

Thus we found that the number of single-digit occurrences of 1 is not only related to single digit, and 10 digits are also related, if single digit is greater than or equal to 1, the number of single digit 1 is 10 digits plus 1, if single digit is 0, the number of single digit 1 is equal to 10 digit number. The number of 10-digit occurrences of 1 is not only related to the 10-digit number, but also to the single-digit: If the 10-digit number equals 1, 10 is the number of digits on the 1-digit number plus 1, and if the 10-digit is greater than 1, 10 is 1 on the 10-digit number.

3-digit situation:

N=123

The number of single digit occurrences is 13:1,11,21,...,91,101,111,121.1

The number of 10-bit occurrences of 1 is 20:10~19,110~119

The number of hundreds of 1 appears to be 24:100~123

We can continue to analyze 4-digit, 5-digit numbers and derive the following general situation:

Suppose N, we are going to calculate the number of occurrences of 1 on the hundred, will be determined by three parts: the number on the hundred, the number above the hundred, the number of the hundred.

If the number on the hundred is 0, then the number of 1 on the hundred is determined only by a higher level, such as 12013, 1 of the case is 100~199,1100~1199,2100~2199,...,11100~11199, a total of 1200. equals the higher number multiplied by the current number of digits, which is 12 * 100.

If the number on the hundred is greater than 1, then the number of occurrences of 1 on the hundred is determined only by higher levels, such as 12213, 1 of the total is 1300 100~199,1100~1199,2100~2199,...,11100~11199,12100~12199. equals the higher number plus 1 times the current number of digits, i.e. (12 + 1) *100.

If the number on the hundred is 1, the number of 1 on the hundred will be affected not only by higher levels but also by low levels. For example 12113, affected by high Level 1 of the situation: 100~199,1100~1199,2100~2199,...,11100~11199, a total of 1200, but it is also affected by the low, 1 of the situation is 12100~12113, a total of 114, equals Low digit 113+1.

To synthesize the above analysis, write the following code:

01	publiclong CountOne2(longn) {

02	longcount = 0;

03	longi = 1;

04	longcurrent = 0, after = 0, before = 0;

05	while((n / i) != 0) {

06	current = (n / i) % 10;

07	before = n / (i * 10);

08	after = n - (n / i) * i;

09

10	if(current > 1)

11	count = count + (before + 1) * i;

12	elseif(current == 0)

13	count = count + before * i;

14	elseif(current == 1)

15	count = count + before * i + after + 1;

16

17	i = i * 10;

18

}

19	returncount;

20

}

Count 1 between 0 and N