字符串减法、ACSII码

AC代码：

普通代码：

import javax.swing.*;
import java.io.BufferedReader;
import java.io.BufferedWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.math.BigInteger;
import java.nio.file.attribute.AclEntryFlag;
import java.security.AlgorithmConstraints;
import java.sql.Struct;
import java.text.CollationElementIterator;
import java.text.DateFormatSymbols;
import java.util.*;
import java.util.stream.Collectors;public class Main
{static PrintWriter pw = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));static int N = (int)2e5 + 10;static math_myself math_me = new math_myself();static int a[] = new int[N];public static void main(String[] args ) throws IOException{char a[] = rd.nextLine().toCharArray();char b[] = rd.nextLine().toCharArray();StringBuilder s = new StringBuilder("");// 将b数组的字符的ascii码都变成1，打个标记int ascii[] = new int [N];for(int i = 0 ; i < b.length ; i ++){ascii[b[i]] = 1;}// 将ascii码不为1的字符输出，即为A-B的结果for(int i = 0 ; i < a.length ; i ++){if(ascii[a[i]] == 1)  continue;else pw.print(a[i]);}pw.println(s);pw.flush();}
}class rd
{static BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));static StringTokenizer tokenizer = new StringTokenizer("");static String nextLine() throws IOException   { return reader.readLine(); }static String next() throws IOException{while (!tokenizer.hasMoreTokens())  tokenizer = new StringTokenizer(reader.readLine());return tokenizer.nextToken();}static int nextInt() throws IOException  { return Integer.parseInt(next()); }static double nextDouble() throws IOException { return Double.parseDouble(next()); }static long nextLong() throws IOException  { return Long.parseLong(next());}static BigInteger nextBigInteger() throws IOException{BigInteger d = new BigInteger(rd.nextLine());return d;}
}class PII
{int x,y;public PII(int x ,int y){this.x = x;this.y = y;}
}class math_myself
{int gcd(int a,int b){if(b == 0)  return a;else return gcd(b,a % b);}int lcm(int a,int b){return a * b / gcd(a, b);}// 求n的所有约数List get_factor(int n){List<Long> a = new ArrayList<>();for(long i = 1; i <= Math.sqrt(n) ; i ++){if(n % i == 0){a.add(i);if(i != n / i)  a.add(n / i);  // // 避免一下的情况：x = 16时，i = 4 ，x / i = 4的情况，这样会加入两种情况  ^-^复杂度能减少多少是多少}}// 相同因子去重，这个方法，完美a = a.stream().distinct().collect(Collectors.toList());// 对因子排序（升序）Collections.sort(a);return a;}// 判断是否是质数boolean check_isPrime(int n){if(n < 2) return false;for(int i = 2 ; i <= n / i; i ++)  if (n % i == 0) return false;return true;}
}

加了StringBuilder的代码：（将A和B中不重叠的字符加到一个新的StringBuilder里面去）

import javax.swing.*;
import java.io.BufferedReader;
import java.io.BufferedWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.math.BigInteger;
import java.nio.file.attribute.AclEntryFlag;
import java.security.AlgorithmConstraints;
import java.sql.Struct;
import java.text.CollationElementIterator;
import java.text.DateFormatSymbols;
import java.util.*;
import java.util.stream.Collectors;public class Main
{static PrintWriter pw = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));static int N = (int)2e5 + 10;static math_myself math_me = new math_myself();static int a[] = new int[N];public static void main(String[] args ) throws IOException{char a[] = rd.nextLine().toCharArray();char b[] = rd.nextLine().toCharArray();StringBuilder s = new StringBuilder("");// 将b数组的字符的ascii码都变成1，打个标记int ascii[] = new int [N];for(int i = 0 ; i < b.length ; i ++){ascii[b[i]] = 1;}// 将ascii码不为1的字符输出，即为A-B的结果for(int i = 0 ; i < a.length ; i ++){if(ascii[a[i]] == 1)  continue;else s.append(a[i]);}pw.println(s);pw.flush();}
}class rd
{static BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));static StringTokenizer tokenizer = new StringTokenizer("");static String nextLine() throws IOException   { return reader.readLine(); }static String next() throws IOException{while (!tokenizer.hasMoreTokens())  tokenizer = new StringTokenizer(reader.readLine());return tokenizer.nextToken();}static int nextInt() throws IOException  { return Integer.parseInt(next()); }static double nextDouble() throws IOException { return Double.parseDouble(next()); }static long nextLong() throws IOException  { return Long.parseLong(next());}static BigInteger nextBigInteger() throws IOException{BigInteger d = new BigInteger(rd.nextLine());return d;}
}class PII
{int x,y;public PII(int x ,int y){this.x = x;this.y = y;}
}class math_myself
{int gcd(int a,int b){if(b == 0)  return a;else return gcd(b,a % b);}int lcm(int a,int b){return a * b / gcd(a, b);}// 求n的所有约数List get_factor(int n){List<Long> a = new ArrayList<>();for(long i = 1; i <= Math.sqrt(n) ; i ++){if(n % i == 0){a.add(i);if(i != n / i)  a.add(n / i);  // // 避免一下的情况：x = 16时，i = 4 ，x / i = 4的情况，这样会加入两种情况  ^-^复杂度能减少多少是多少}}// 相同因子去重，这个方法，完美a = a.stream().distinct().collect(Collectors.toList());// 对因子排序（升序）Collections.sort(a);return a;}// 判断是否是质数boolean check_isPrime(int n){if(n < 2) return false;for(int i = 2 ; i <= n / i; i ++)  if (n % i == 0) return false;return true;}
}

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