1~n整数中1出现的次数

package array_2_3_1;
import java.util.Scanner;
/*** 输入整数n，求1-n中，1出现的次数* 思路：* 1.从个位开始遍历,直到最高位* 2.“1”出现的次数为：每个位上，组合high mid(求指向该位数时，1出现的组合次数) low*/
public class Demo_43 {public static void main(String[] args) {Scanner sc = new Scanner(System.in);int n = sc.nextInt();Demo_43 demo = new Demo_43();int res = demo.calculate1(n);System.out.println(res);}private int calculate1(int n) {int totalWei = (n+"").length();//计算整数n的位数int num = 0;//1在每个数位上出现的次数int sum = 0;//1总共出现的次数int high = 0;//高位数int mid = 0; //数值本身int low = 0;//低位数for (int i = 0,step = 1; i < totalWei; i++,step *= 10) {high = n / (step * 10);mid = n / step % 10;low = n % step;if (mid == 0){num = (int) ((high - 1 + 1) * Math.pow(10,i));//向前面一位借1}else if (mid > 1){num = (int) ((high + 1) * Math.pow(10,i ));}else {num = (int) (high * Math.pow(10,i ) + (low+1));}sum += num;}return sum;}
}