在Java中有一种方法来检查条件:
"这个字符是否出现在字符串x中"
不使用循环?
在Java中有一种方法来检查条件:
"这个字符是否出现在字符串x中"
不使用循环?
当前回答
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
其他回答
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
String temp = "abcdefghi";
if(temp.indexOf("b")!=-1)
{
System.out.println("there is 'b' in temp string");
}
else
{
System.out.println("there is no 'b' in temp string");
}
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))。
如果您需要经常检查相同的字符串,您可以预先计算字符出现的次数。这是一个使用位数组包含在长数组中的实现:
public class FastCharacterInStringChecker implements Serializable {
private static final long serialVersionUID = 1L;
private final long[] l = new long[1024]; // 65536 / 64 = 1024
public FastCharacterInStringChecker(final String string) {
for (final char c: string.toCharArray()) {
final int index = c >> 6;
final int value = c - (index << 6);
l[index] |= 1L << value;
}
}
public boolean contains(final char c) {
final int index = c >> 6; // c / 64
final int value = c - (index << 6); // c - (index * 64)
return (l[index] & (1L << value)) != 0;
}}
package com;
public class _index {
public static void main(String[] args) {
String s1="be proud to be an indian";
char ch=s1.charAt(s1.indexOf('e'));
int count = 0;
for(int i=0;i<s1.length();i++) {
if(s1.charAt(i)=='e'){
System.out.println("number of E:=="+ch);
count++;
}
}
System.out.println("Total count of E:=="+count);
}
}