2019互聯網筆試題編程答案部分

阿里

import java.util.Scanner;

public class Main{
    public static void main(String args[])
    {
        Scanner sc=new Scanner(System.in);
        int n=sc.nextInt();
        int result=getResult(n-1,n)+getResult(n-2,n)+getResult(n-3,n);
        System.out.println(result);
    }
    public static int getResult(int n,int m)
    {
        if(n==0) return 1;
        if(n==1 || n==2) return 1;
        if(n==m-1)
        {
            return getResult(n-2,n)+getResult(n-3,n);
        }
        else
        {
            return getResult(n-1,n)+getResult(n-2,n)+getResult(n-3,n);
        }
    }
}

網易有道

import java.util.Scanner;
public class Main {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int t = in.nextInt();
        for(int i = 0; i < t; i++) {
            int min = 0, max;
            int n = in.nextInt();
            int k = in.nextInt();
            if(n % 2 == 0) {
                if(k > n / 2) {
                    max = n - k;
                } else {
                    max = k - 1;
                }
            } else {
                if(k > n / 2 + 1) {
                    max = n - k;
                } else {
                    max = k - 1;
                }
            }
            if(n < 3) {
                max = 0;
            }
            if(max < 0) {
                max = 0;
            }
            System.out.println(min + " " + max);
        }
    }
}

頭條第三次筆試

1，給定字符串輸出字符串中不重複的最長子串的長度 例如：輸入‘acddw’ 輸出3  輸入‘bbbb’ 輸出 1

def solution(one_str):
    res_list=[]
    length=len(one_str)
    for i in range(length):
        tmp=one_str[i]
        for j in range(i+1, length):
            if one_str[j] not in tmp:
                tmp+=one_str[j]
            else:
                break
        res_list.append(tmp)
    res_list.sort(lambda x,y:cmp(len(x),len(y)))
    return res_list[-1]
ss=solution(s)
print len(ss)