L1-011 A-B (20分)

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.Scanner;

public class Main{
    public static String fun(String a, String b){
        char[] arrA = a.toCharArray();
        char[] arrB = b.toCharArray();
        ArrayList<Character> listA = new ArrayList<>();
        ArrayList<Character> listB = new ArrayList<>();
        int len1 = arrA.length;
        int len2 = arrB.length;
        for (int i = 0; i < len1; i++){
            listA.add(arrA[i]);
        }
        for (int j = 0; j < len2; j++){
            listB.add(arrB[j]);
        }
        listA.removeAll(listB);
        Character[] ans = new Character[listA.size()];
        listA.toArray(ans);
        int len3 = ans.length;
        char[] newAns = new char[len3];
        for (int i = 0; i < len3; i++){
            newAns[i] = ans[i];
        }
        String rtn = new String(newAns);
        return rtn;

    }

    public static void main(String[] args){
//        Scanner input = new Scanner(System.in);
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        String a = null;
        String b = null;
        try {
             a = br.readLine();
        } catch (IOException e) {
            e.printStackTrace();
        }
        try {
             b = br.readLine();
        } catch (IOException e) {
            e.printStackTrace();
        }



        System.out.println(Main.fun(a,b));
    }
}

我就知道第一次会超时

然后先把数组长度算出来反而更加耗时了

​​看这位前辈说是用Java过不去​​看起来这个是Java里面最优解，但还是有一个点过不去

import java.io.*;
public class Main{
  public static void main(String[] args) throws IOException {
    BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
      String  A=in.readLine();
      String  B=in.readLine();
      System.out.print(A.replaceAll("[" + B + "]", "")); 
      }
    }

终于过去了
解法：辣么多ASCII码也在0-255之间，所以用数组标记所有的ASCII码,如果第二个字符出现了这个ACSII码那就标记为1~然后输出的时候当数组对应的那个ASCII为1的时候就跳过不输出

#include<stdio.h>

char a[10001];
char b[10001];
char c[256];
int main() // 推荐
{
  gets(a);
  gets(b);
  int len = strlen(a);
  int len1 = strlen(b);
  for (int i = 0; i < len1; ++i)
  {
    c[b[i]] = 1;
  }
  for (int i = 0; i < len; ++i)
  {
    if (!c[a[i]])  putchar(a[i]);
  }
  putchar('\n');
  return 0;
}