题解 | #求最大连续bit数#

import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        String str = intToStr2(n);
        int max = getLength(str);
        System.out.println(max);
    }

    private static int getLength(String str) {
        int max = 0;
        int length = str.length();
        for (int i = 0; i < length; i++) {
            if (str.charAt(i) == '1') {
                int count = 0;
                for (int j = i; j < length; j++) {
                    if (str.charAt(j) == '1') {
                        count++;
                    } else {
                        break;
                    }
                }
                max = Math.max(max, count);
            }
        }
        return max;
    }

    private static String intToStr2(int n) {
        String s = "";
        while (n != 0) {
            int yushu = n % 2;
            s = yushu + s;
            n = n / 2;
        }
        return s;
    }
}

解题思路:

1, 先将整数转化为二进制字符串,

2, 再对二进制字符串进行遍历, 就能找到最长的字符串.