Java implementation to find a number within a range of all

I. Contents of the topic

Given a positive integer in decimal, write down all integers starting with 1, to N, and then count the number of "1" that appears.
Requirements:
Write a function f (n) that returns the number of "1" that occurs between 1 and N. For example F (12) = 5.
Within a 32-bit integer range, the maximum n of the "F (n) =n" that satisfies the condition is what.

Second, design ideas

Each bit to calculate the number of 1, each adjacent to the three-digit link, ABC, first count C, plus B, and finally plus a

Third, the experimental code

Findone.java Package Com.minirisoft;import Java.util.scanner;public class FindOne {public static int Sum (int n) {         int Count = 0;         int Factor = 1;         int Lower = 0;         int Curr = 0;         int higher = 0;           while (N/factor! = 0) {Lower = n-(n/factor) * Factor;           Curr = (n/factor)% 10;                 Higher = N/(Factor *10);                  Switch (Curr) {case 0:count + = higher * Factor;              Break                  Case 1:count + = higher * Factor + Lower + 1;              Break                  Default:count + = (higher + 1) * Factor;              Break         } Factor *= 10;      } return Count;        } public static void Main (string[] args) {Scanner input=new Scanner (system.in);        System.out.print ("Please enter a positive integer:");    int N=input.nextint (); System.out.print ("1 to" +n+ "1 numbers: "+sum (N) +" x "); }   }

Four, the experiment

Five, experience

General design procedures, to find the law of the problem, according to the law in the process of making

Java implementation to find a number within a range of all