you can use this code. It will check the char is present or not. If it is present then the return value is >= 0 otherwise it's -1. Here I am printing alphabets that is not present in the input.

import java.util.Scanner;

public class Test {

public static void letters()
{
    System.out.println("Enter input char");
    Scanner sc = new Scanner(System.in);
    String input = sc.next();
    System.out.println("Output : ");
    for (char alphabet = 'A'; alphabet <= 'Z'; alphabet++) {
            if(input.toUpperCase().indexOf(alphabet) < 0) 
                System.out.print(alphabet + " ");
    }
}
public static void main(String[] args) {
    letters();
}

}

//Ouput Example
Enter input char
nandu
Output : 
B C E F G H I J K L M O P Q R S T V W X Y Z

是的，在字符串类上使用indexOf()方法。请参阅此方法的API文档

您可以使用String类中的两个方法。

string .contains()检查字符串是否包含指定的char值序列 string . indexof()返回字符串中第一次出现指定字符或子字符串的索引，如果没有找到该字符则返回-1(此方法有4种变体)

方法1:

String myString = "foobar";
if (myString.contains("x") {
    // Do something.
}

方法2:

String myString = "foobar";
if (myString.indexOf("x") >= 0 {
    // Do something.
}

链接作者:Zach Scrivena

string .contains()检查字符串是否包含指定的char值序列 string . indexof()返回字符串中第一次出现指定字符或子字符串的索引(此方法有4种变体)

you can use this code. It will check the char is present or not. If it is present then the return value is >= 0 otherwise it's -1. Here I am printing alphabets that is not present in the input.

import java.util.Scanner;

public class Test {

public static void letters()
{
    System.out.println("Enter input char");
    Scanner sc = new Scanner(System.in);
    String input = sc.next();
    System.out.println("Output : ");
    for (char alphabet = 'A'; alphabet <= 'Z'; alphabet++) {
            if(input.toUpperCase().indexOf(alphabet) < 0) 
                System.out.print(alphabet + " ");
    }
}
public static void main(String[] args) {
    letters();
}

}

//Ouput Example
Enter input char
nandu
Output : 
B C E F G H I J K L M O P Q R S T V W X Y Z

如果不使用循环/递归至少检查一次字符串，您将无法检查char是否出现在某些字符串中(像indexOf这样的内置方法也使用循环)

如果不是。如果你在字符串中查找一个字符，x比字符串的长度要多得多，我建议使用Set数据结构，因为这比简单地使用indexOf更有效

String s = "abc";

// Build a set so we can check if character exists in constant time O(1)
Set<Character> set = new HashSet<>();
int len = s.length();
for(int i = 0; i < len; i++) set.add(s.charAt(i));

// Now we can check without the need of a loop
// contains method of set doesn't use a loop unlike string's contains method
set.contains('a') // true
set.contains('z') // false

使用set，你将能够在常数时间O(1)检查字符是否存在于字符串中，但你也将使用额外的内存(空间复杂度将是O(n))。