题解 | #求最大连续bit数#--Integer方法及排序

import java.util.*;

public class Main {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        while (in.hasNextInt()) {
            int num = in.nextInt();
            // 转化为二进制字符串
            String binaryStr = Integer.toBinaryString(num);
            int count = Integer.bitCount(num);
            if (count == 1) {
                System.out.println(1);
            } else if (count == 0) {
                System.out.println(0);
            } else {
                // 统计所有1连续出现的次数
                int consistCount = 0;
                List<Integer> countList = new ArrayList<>();
                for (int i = 0; i < binaryStr.length(); i++) {
                    if (binaryStr.charAt(i) == '1') {
                        consistCount++;
                        countList.add(consistCount);
                    } else {
                        consistCount = 0;
                    }
                }
                // 输出1最大连续出现次数
                System.out.println(countList.stream().max(Integer :: compareTo).get());
            }
        }
    }
}