求一个数内含1个次数

import java.util.regex.Matcher;
import java.util.regex.Pattern;

/****

****author :hxy

****

****/

public class Test {

    private final int n;
    public int cnt_all;

    public Test(int n) {
        this.n = n;
        cnt_all = 0;
    }

    // 分多个线程计算
    public void binaryHandlerN(int threadNum) {
        int gap = n / threadNum;
        int tmp = 0;
        do {
            Thread t = new Thread(new _handler(tmp + 1, tmp + gap));
            tmp = tmp + gap;
            t.run();
        } while (tmp < n);
    }

    // 匹配每个数有几个1
    private int rexStr(int i) {
        int cntForEachNum = 0;
        Matcher m = Pattern.compile(String.valueOf(1)).matcher(String.valueOf(i));
        while (m.find()) {
            cntForEachNum++;
        }
        return cntForEachNum;
    }

    class _handler implements Runnable {
        int cnt = 0;
        int num1 = 0;
        int num2 = 0;

        public _handler(int num1, int num2) {
            this.num1 = num1;
            this.num2 = num2;
            if (num2 > n) {
                this.num2 = n;
            }
        }

        @Override
        public void run() {
            int cnt = 0;
            for (int i = num1; i <= num2; i++) {
                cnt = cnt + rexStr(i);
            }
            cnt_all = cnt + cnt_all;
        }
    }

    public static void main(String[] args) {
        Test t = new Test(44);
        t.binaryHandlerN(5);
        System.out.println("1重复的次数为：" + t.cnt_all);
    }
}