blocks|key|351277|text|您可以使用string.indexOf('a')。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|351278|如果string中存在char+a：|351279|351280|它返回该对象表示的字符序列中该字符第一次出现的索引，如果该字符没有出现，则返回-1。|blockquote|351281|351282|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.oracle.com/javase/7/docs/api/java/lang/String.html#indexOf%2528int%2529^0|5|J|5|J|0|0|2|6|G|1|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@$9|V|A|W|B|C]]|D|@$9|X|A|Y|1|Z]]|E|$]]|$1|F|3|G|5|6|7|10|8|@$9|11|A|12|B|C]|$9|13|A|14|B|C]]|D|@]|E|$]]|$1|H|3|-4|5|6|7|15|8|@]|D|@]|E|$]]|$1|I|3|J|5|K|7|16|8|@]|D|@]|E|$]]|$1|L|3|-4|5|6|7|17|8|@]|D|@]|E|$]]|$1|M|3|-4|5|6|7|18|8|@]|D|@]|E|$]]]|N|$O|$5|P|Q|R|E|$S|T]]]]

You can use <a href="http://docs.oracle.com/javase/7/docs/api/java/lang/String.html#indexOf%28int%29" rel="noreferrer"><code>string.indexOf('a')</code></a>.

If the char <code>a</code> is present in <code>string</code> : 

<blockquote>
 it returns the the index of the first occurrence of the character in
 the character sequence represented by this object, or -1 if the
 character does not occur.
</blockquote>

blocks|key|3343371|text|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3343372|检查字符串是否包含指定的字符序列values|unordered-list-item|3343373|返回指定字符或子字符串首次出现的字符串内的索引(此方法有4种变体)|3343374|3343375|entityMap^0|0|0|0|0^^$0|@$1|2|3|-4|4|5|6|I|7|@]|8|@]|9|$]]|$1|A|3|B|4|C|6|J|7|@]|8|@]|9|$]]|$1|D|3|E|4|C|6|K|7|@]|8|@]|9|$]]|$1|F|3|-4|4|5|6|L|7|@]|8|@]|9|$]]|$1|G|3|-4|4|5|6|M|7|@]|8|@]|9|$]]]|H|$]]

<ul>
<li><a href="http://java.sun.com/javase/6/docs/api/java/lang/String.html#contains(java.lang.CharSequence)" rel="noreferrer"><code>String.contains()</code></a> which checks if the string contains a specified sequence of char values</li>
<li><a href="http://java.sun.com/javase/6/docs/api/java/lang/String.html#indexOf(int)" rel="noreferrer"><code>String.indexOf()</code></a> which returns the index within the string of the first occurence of the specified character or substring (there are 4 variations of this method)</li>
</ul>

blocks|key|3343396|text|要检查字符串中是否不存在某些内容，至少需要查看字符串中的每个字符。因此，即使您不显式地使用循环，它也将具有相同的效率。也就是说，您可以尝试使用str.contains(""%2Bchar)。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3343397|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

To check if something does not exist in a string, you at least need to look at each character in a string. So even if you don't explicitly use a loop, it'll have the same efficiency. That being said, you can try using str.contains(""+char).

blocks|key|354054|text|是的，在string类上使用indexOf()方法。See+the+API+documentation+for+this+method|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|354055|entityMap|0|LINK|mutability|MUTABLE|url|http://java.sun.com/j2se/1.5.0/docs/api/java/lang/String.html#indexOf(int)^0|Q|15|0|0^^$0|@$1|2|3|4|5|6|7|L|8|@]|9|@$A|M|B|N|1|O]]|C|$]]|$1|D|3|-4|5|6|7|P|8|@]|9|@]|C|$]]]|E|$F|$5|G|H|I|C|$J|K]]]]

Yes, using the indexOf() method on the string class. <a href="http://java.sun.com/j2se/1.5.0/docs/api/java/lang/String.html#indexOf(int)" rel="nofollow noreferrer">See the API documentation for this method</a>

blocks|key|349011|text|我不确定最初的发帖者到底在问什么。由于indexOf(...)并包含(...)两者都可能在内部使用循环，也许他想看看在没有循环的情况下这是否可能？我可以随手想到两种方法，一种当然是递归：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|349012|public+boolean+containsChar(String+s,+char+search)+{
++++if+(s.length()+==+0)
++++++++return+false;
++++else
++++++++return+s.charAt(0)+==+search+%7C%7C+containsChar(s.substring(1),+search);
}|code-block|syntax|javascript|349013|另一个远没有那么优雅，但完备性...：|349014|/**
+*+Works+for+strings+of+up+to+5+characters
+*/
public+boolean+containsChar(String+s,+char+search)+{
++++if+(s.length()+>+5)+throw+IllegalArgumentException();

++++try+{
++++++++if+(s.charAt(0)+==+search)+return+true;
++++++++if+(s.charAt(1)+==+search)+return+true;
++++++++if+(s.charAt(2)+==+search)+return+true;
++++++++if+(s.charAt(3)+==+search)+return+true;
++++++++if+(s.charAt(4)+==+search)+return+true;
++++}+catch+(IndexOutOfBoundsException+e)+{
++++++++//+this+should+never+happen...
++++++++return+false;
++++}
++++return+false;
}|349015|当然，当您需要支持越来越长的字符串时，行的数量也会增加。但是根本没有循环/递归。如果您担心length()使用循环，您甚至可以删除长度检查。|349016|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|S|8|@]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

I'm not sure what the original poster is asking exactly. Since indexOf(...) and contains(...) both probably use loops internally, perhaps he's looking to see if this is possible at all without a loop? I can think of two ways off hand, one would of course be recurrsion:

<pre><code>public boolean containsChar(String s, char search) {
 if (s.length() == 0)
 return false;
 else
 return s.charAt(0) == search || containsChar(s.substring(1), search);
}
</code></pre>

The other is far less elegant, but completeness...:

<pre><code>/**
 * Works for strings of up to 5 characters
 */
public boolean containsChar(String s, char search) {
 if (s.length() &gt; 5) throw IllegalArgumentException();

 try {
 if (s.charAt(0) == search) return true;
 if (s.charAt(1) == search) return true;
 if (s.charAt(2) == search) return true;
 if (s.charAt(3) == search) return true;
 if (s.charAt(4) == search) return true;
 } catch (IndexOutOfBoundsException e) {
 // this should never happen...
 return false;
 }
 return false;
}
</code></pre>

The number of lines grow as you need to support longer and longer strings of course. But there are no loops/recurrsions at all. You can even remove the length check if you're concerned that that length() uses a loop.

blocks|key|3130604|text|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");
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|3130605|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>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");
}
</code></pre>

blocks|key|351315|text|如果你需要经常检查同一个字符串，你可以预先计算字符出现的次数。这是一个使用包含在长数组中的位数组的实现：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|351316|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]+%7C=+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;
}}|code-block|syntax|javascript|351317|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

If you need to check the same string often you can calculate the character occurrences up-front. This is an implementation that uses a bit array contained into a long array:

<pre><code>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 &gt;&gt; 6;
 final int value = c - (index &lt;&lt; 6);
 l[index] |= 1L &lt;&lt; value;
 }
}

public boolean contains(final char c) {
 final int index = c &gt;&gt; 6; // c / 64
 final int value = c - (index &lt;&lt; 6); // c - (index * 64)
 return (l[index] &amp; (1L &lt;&lt; value)) != 0;
}}
</code></pre>

blocks|key|3130614|text|static+String+removeOccurences(String+a,+String+b)
{
++++StringBuilder+s2+=+new+StringBuilder(a);

++++for(int+i=0;i<b.length();i%2B%2B){
++++++++char+ch+=+b.charAt(i);++
++++++++System.out.println(ch%2B"++first+index"%2Ba.indexOf(ch));

++++++++int+lastind+=+a.lastIndexOf(ch);

++++for(int+k=new+String(s2).indexOf(ch);k+>+0;k=new+String(s2).indexOf(ch)){
++++++++++++if(s2.charAt(k)+==+ch){
++++++++++++++++s2.deleteCharAt(k);
++++++++System.out.println("val+of+s2+:+++++++++++++"%2Bs2.toString());
++++++++++++}
++++++++}
++++++}

++++System.out.println(s1.toString());

++++return+(s1.toString());
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|3130615|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>static String removeOccurences(String a, String b)
{
 StringBuilder s2 = new StringBuilder(a);

 for(int i=0;i&lt;b.length();i++){
 char ch = b.charAt(i); 
 System.out.println(ch+" first index"+a.indexOf(ch));

 int lastind = a.lastIndexOf(ch);

 for(int k=new String(s2).indexOf(ch);k &gt; 0;k=new String(s2).indexOf(ch)){
 if(s2.charAt(k) == ch){
 s2.deleteCharAt(k);
 System.out.println("val of s2 : "+s2.toString());
 }
 }
 }

 System.out.println(s1.toString());

 return (s1.toString());
}
</code></pre>

blocks|key|349033|text|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%2B%2B)+{
++++++++++++if(s1.charAt(i)=='e'){
++++++++++++++++System.out.println("number+of+E:=="%2Bch);
++++++++++++++++count%2B%2B;
++++++++++++}
++++++++}
++++++++System.out.println("Total+count+of+E:=="%2Bcount);
++++}
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|349034|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>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&lt;s1.length();i++) {
 if(s1.charAt(i)=='e'){
 System.out.println("number of E:=="+ch);
 count++;
 }
 }
 System.out.println("Total count of E:=="+count);
 }
}
</code></pre>

blocks|key|3343472|text|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%2B%2B)+{
++++++++++++if(input.toUpperCase().indexOf(alphabet)+<+0)+
++++++++++++++++System.out.print(alphabet+%2B+"+");
++++}
}
public+static+void+main(String[]+args)+{
++++letters();
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|3343473|}|unstyled|3343474|//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|3343475|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|L|8|@]|9|@]|A|$]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$B|C]]|$1|I|3|-4|5|F|7|N|8|@]|9|@]|A|$]]]|J|$]]

<pre><code>you can use this code. It will check the char is present or not. If it is present then the return value is &gt;= 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 &lt;= 'Z'; alphabet++) {
 if(input.toUpperCase().indexOf(alphabet) &lt; 0) 
 System.out.print(alphabet + " ");
 }
}
public static void main(String[] args) {
 letters();
}
</code></pre>

}

<pre><code>//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
</code></pre>

blocks|key|354079|text|您可以使用String类中的两个方法。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|354080|354081|检查字符串是否包含指定的字符序列values|unordered-list-item|354082|返回指定字符或子串第一次出现的字符串中的索引，如果未找到该字符则返回-1+(此方法有4种变体)|354083|354084|方法1：|354085|String+myString+=+"foobar";
if+(myString.contains("x")+{
++++//+Do+something.
}|code-block|syntax|javascript|354086|方法二：|354087|String+myString+=+"foobar";
if+(myString.indexOf("x")+>=+0+{
++++//+Do+something.
}|354088|链接者：Zach+Scrivena|354089|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/users/20029/zach-scrivena^0|5|6|0|0|0|0|0|0|0|0|0|4|D|0|0^^$0|@$1|2|3|4|5|6|7|17|8|@$9|18|A|19|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|1A|8|@]|D|@]|E|$]]|$1|G|3|H|5|I|7|1B|8|@]|D|@]|E|$]]|$1|J|3|K|5|I|7|1C|8|@]|D|@]|E|$]]|$1|L|3|-4|5|6|7|1D|8|@]|D|@]|E|$]]|$1|M|3|N|5|6|7|1E|8|@]|D|@]|E|$]]|$1|O|3|P|5|Q|7|1F|8|@]|D|@]|E|$R|S]]|$1|T|3|U|5|6|7|1G|8|@]|D|@]|E|$]]|$1|V|3|W|5|Q|7|1H|8|@]|D|@]|E|$R|S]]|$1|X|3|Y|5|6|7|1I|8|@]|D|@$9|1J|A|1K|1|1L]]|E|$]]|$1|Z|3|-4|5|6|7|1M|8|@]|D|@]|E|$]]]|10|$11|$5|12|13|14|E|$15|16]]]]

You can use 2 methods from the <code>String</code> class.

<ul>
<li><a href="http://java.sun.com/javase/6/docs/api/java/lang/String.html#contains(java.lang.CharSequence)" rel="noreferrer"><code>String.contains()</code></a> which checks if the string contains a specified sequence of char values</li>
<li><a href="http://java.sun.com/javase/6/docs/api/java/lang/String.html#indexOf(int)" rel="noreferrer"><code>String.indexOf()</code></a> which returns the index within the string of the first occurence of the specified character or substring or returns -1 if the character is not found (there are 4 variations of this method)</li>
</ul>

Method 1:

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

Method 2:

<pre><code>String myString = "foobar";
if (myString.indexOf("x") &gt;= 0 {
 // Do something.
}
</code></pre>

Links by: <a href="https://stackoverflow.com/users/20029/zach-scrivena">Zach Scrivena</a>

blocks|key|354091|text|下面是你要找的吗？|type|unstyled|depth|inlineStyleRanges|entityRanges|data|354092|int+index+=+string.indexOf(character);
return+index+!=+-1;|code-block|syntax|javascript|354093|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Is the below what you were looking for?
<pre><code>int index = string.indexOf(character);
return index != -1;
</code></pre>

blocks|key|350343|text|如果不使用循环/递归检查字符串一次，您将无法检查字符是否出现在某个字符串中(内置方法，如indexOf也使用循环)|type|unstyled|depth|inlineStyleRanges|entityRanges|data|350344|如果不是。如果字符在字符串中，查找的次数x比我建议使用Set数据结构要多得多，因为这比简单地使用indexOf更有效|offset|length|style|CODE|350345|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%2B%2B)+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|code-block|syntax|javascript|350346|使用set，您将能够在constant+time+O(1)中检查字符串中是否存在字符，但您还将使用额外的内存(空间复杂度将为O(n)+)。|BOLD|350347|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.oracle.com/javase/7/docs/api/java/util/Set.html^0|0|1C|7|R|3|0|0|0|B|D|0^^$0|@$1|2|3|4|5|6|7|X|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|Y|8|@$D|Z|E|10|F|G]]|9|@$D|11|E|12|1|13]]|A|$]]|$1|H|3|I|5|J|7|14|8|@]|9|@]|A|$K|L]]|$1|M|3|N|5|6|7|15|8|@$D|16|E|17|F|O]]|9|@]|A|$]]|$1|P|3|-4|5|6|7|18|8|@]|9|@]|A|$]]]|Q|$R|$5|S|T|U|A|$V|W]]]]

You won't be able to check if char appears at all in some string without atleast going over the string once using loop / recursion ( the built-in methods like indexOf also use a loop )

If the no. of times you look up if a char is in string x is more way more than the length of the string than I would recommend using a <a href="https://docs.oracle.com/javase/7/docs/api/java/util/Set.html" rel="nofollow noreferrer">Set</a> data structure as that would be more efficient than simply using <code>indexOf</code>

<pre><code>String s = "abc";

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

Using set you will be able to check if character exists in a string in constant time O(1) but you will also use additional memory ( Space complexity will be O(n) ).

blocks|key|3130725|text|如果您在JAVA中看到indexOf的源代码：|type|unstyled|depth|inlineStyleRanges|offset|length|style|BOLD|entityRanges|data|3130726|public+int+indexOf(int+ch,+int+fromIndex)+{

++++++++final+int+max+=+value.length;

++++++++if+(fromIndex+<+0)+{

++++++++++++fromIndex+=+0;

++++++++}+else+if+(fromIndex+>=+max)+{

++++++++++++//+Note:+fromIndex+might+be+near+-1>>>1.

++++++++++++return+-1;

++++++++}


++++++++if+(ch+<+Character.MIN_SUPPLEMENTARY_CODE_POINT)+{

++++++++++++//+handle+most+cases+here+(ch+is+a+BMP+code+point+or+a

++++++++++++//+negative+value+(invalid+code+point))

++++++++++++final+char[]+value+=+this.value;

++++++++++++for+(int+i+=+fromIndex;+i+<+max;+i%2B%2B)+{

++++++++++++++++if+(value[i]+==+ch)+{

++++++++++++++++++++return+i;

++++++++++++++++}

++++++++++++}

++++++++++++return+-1;

++++++++}+else+{

++++++++++++return+indexOfSupplementary(ch,+fromIndex);

++++++++}

++++}|code-block|syntax|javascript|3130727|您可以看到它使用了一个for循环来查找字符。请注意，您可以在代码中使用的每个indexOf都等于一个循环。|3130728|因此，对单个字符使用是不可避免的。|3130729|但是，如果您想查找具有更多不同形式的特殊字符串，请使用有用的库，如util.regex，它部署了更强大的算法来匹配字符或字符串模式与正则表达式。例如，要在字符串中查找电子邮件：|CODE|3130730|String+regex+=+"%5E(.%2B)@(.%2B)$";
+
Pattern+pattern+=+Pattern.compile(regex);
Matcher+matcher+=+pattern.matcher(email);|3130731|如果您不喜欢使用正则表达式，只需使用循环和charAt，并尝试在一个循环中涵盖所有情况。|3130732|要小心，递归方法比循环方法有更多的开销，所以不推荐这样做。|3130733|entityMap^0|B|7|0|0|0|A|7|0|X|A|0|0|L|6|0|0^^$0|@$1|2|3|4|5|6|7|Z|8|@$9|10|A|11|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|12|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|13|8|@]|D|@]|E|$]]|$1|M|3|N|5|6|7|14|8|@$9|15|A|16|B|C]]|D|@]|E|$]]|$1|O|3|P|5|6|7|17|8|@$9|18|A|19|B|Q]]|D|@]|E|$]]|$1|R|3|S|5|H|7|1A|8|@]|D|@]|E|$I|J]]|$1|T|3|U|5|6|7|1B|8|@$9|1C|A|1D|B|Q]]|D|@]|E|$]]|$1|V|3|W|5|6|7|1E|8|@]|D|@]|E|$]]|$1|X|3|-4|5|6|7|1F|8|@]|D|@]|E|$]]]|Y|$]]

If you see the source code of indexOf in JAVA:
<pre><code>public int indexOf(int ch, int fromIndex) {

 final int max = value.length;

 if (fromIndex &lt; 0) {

 fromIndex = 0;

 } else if (fromIndex &gt;= max) {

 // Note: fromIndex might be near -1&gt;&gt;&gt;1.

 return -1;

 }


 if (ch &lt; Character.MIN_SUPPLEMENTARY_CODE_POINT) {

 // handle most cases here (ch is a BMP code point or a

 // negative value (invalid code point))

 final char[] value = this.value;

 for (int i = fromIndex; i &lt; max; i++) {

 if (value[i] == ch) {

 return i;

 }

 }

 return -1;

 } else {

 return indexOfSupplementary(ch, fromIndex);

 }

 }
</code></pre>
you can see it uses a for loop for finding a character. Note that each indexOf you may use in your code, is equal to one loop.
So, it is unavoidable to use loop for a single character.
However, if you want to find a special string with more different forms, use useful libraries such as <code>util.regex</code>, it deploys stronger algorithm to match a character or a string pattern with Regular Expressions. For example to find an email in a string:
<pre><code>String regex = &quot;^(.+)@(.+)$&quot;;
 
Pattern pattern = Pattern.compile(regex);
Matcher matcher = pattern.matcher(email);
</code></pre>
If you don't like to use regex, just use a loop and <code>charAt</code> and try to cover all cases in one loop.
Be careful recursive methods has more overhead than loop, so it's not recommended.

blocks|key|3343603|text|String.contains(String)或String.indexOf(String)+-建议|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3343604|"abc".contains("Z");+//+false+-+correct
"zzzz".contains("Z");+//+false+-+correct
"Z".contains("Z");+//+true+-+correct
"and".contains("");+//+true+-+correct
"and".contains("");+//+false+-+correct
"and".indexOf("");+//+0+-+correct
"and".indexOf("");+//+-1+-+correct|code-block|syntax|javascript|3343605|String.indexOf(int)，并仔细考虑使用字符到int扩展的String.indexOf(char)|3343606|"and".indexOf("".charAt(0));+//+0+though+incorrect+usage+has+correct+output+due+to+portion+of+correct+data
"and".indexOf("".charAt(0));+//+0+--+incorrect+usage+and+ambiguous+result
"and".indexOf("".codePointAt(0));+//+-1+--+correct+usage+and+correct+output|3343607|在Java世界中，围绕字符的讨论是模棱两可的|3343608|char或Character的值可以视为单个字符吗？|3343609|No。在unicode字符的上下文中，char或Character有时可以是part+of+a+single+character，在逻辑上不应被视为a+complete+single+character。|BOLD|3343610|如果不是，什么应该被认为是单字符(逻辑上)？|3343611|任何支持Unicode字符的字符编码的系统都应该将unicode的代码点视为单个字符。|3343612|所以Java应该做得非常清楚和响亮，而不是向用户公开太多的内部实现细节。|3343613|String类不擅长抽象(尽管它需要大量令人困惑的understanding+of+its+encapsulations+to+understand+the+abstraction，因此需要一个anti-pattern)。|3343614|它与一般的char用法有什么不同？|3343615|char只能映射到基本多语种平面中的字符。|3343616|只有codePoint+-+int可以覆盖完整的Unicode字符范围。|3343617|为什么会有这种不同呢？|3343618|char在内部被视为16-bit无符号值，并且不能使用仅使用2-bytes的UTF-16内部表示来表示所有的unicode字符。有时，16-bit范围中的值必须与另一个16-bit值组合才能正确定义字符。|3343619|在不太冗长的情况下，indexOf、charAt、length和这类方法的用法应该更加明确。真诚地希望Java能够添加新的具有明确定义的抽象的UnicodeString和UnicodeCharacter类。|3343620|选择contains而不是indexOf(int)的理由|3343621|实际上，在java中有许多将逻辑字符视为char的代码流。在Unicode上下文中，|3343622|+char不是indexOf接受的int，char到<代码>D44的转换对用户屏蔽了这一点，用户可能会做一些类似<代码>D45(除非用户知道确切的上下文)<代码>H246<代码>H147的事情，因此，将一切都视为<代码>D48(也称为<代码>D49)更好<代码>H250<代码>G251|unordered-list-item|3343623|3343624|++++public+static+void+main(String[]+args)+{
++++++++System.out.println("and".indexOf("".charAt(0)));+//+0+though+incorrect+usage+has+correct+output+due+to+portion+of+correct+data
++++++++System.out.println("and".indexOf("".charAt(0)));+//+0+--+incorrect+usage+and+ambiguous+result
++++++++System.out.println("and".indexOf("".codePointAt(0)));+//+-1+--+correct+usage+and+correct+output
++++++++System.out.println("and".contains(""));+//+true+-+correct
++++++++System.out.println("and".contains(""));+//+false+-+correct
++++}|3343625|语义学|3343626|3343627|char可以处理大多数实际用例。尽管如此，在未来的编程环境中使用代码点更好，extensibility.|ordered-list-item|3343628|codepoint应该处理几乎所有围绕encodings.|3343629|Still，的技术用例，abstraction.|3343630|Storage层的Grapheme+Clusters不属于codepoint级别的范围，如果char太昂贵(加倍)，可以选择int接口。除非存储成本是唯一的指标，否则还是使用codepoint更好。此外，最好将存储视为byte，并将语义委托给围绕storage.|3343631|Semantics构建的业务逻辑，可以在多个级别上对其进行抽象。codepoint应该成为最低级别的接口，并且可以在运行时environment.|3343632|中围绕codepoint构建其他语义|3343633|entityMap|0|LINK|mutability|MUTABLE|url|https://unicode.org/reports/tr29/#Grapheme_Cluster_Boundaries^0|0|N|O|M|0|0|0|J|U|3|10|K|0|0|0|0|4|5|9|0|0|2|J|4|O|9|12|Q|22|R|0|0|0|0|0|6|P|1T|2P|C|0|5|4|0|0|4|0|2|F|0|0|0|4|A|6|U|7|1V|6|2C|6|0|A|7|I|6|P|6|1Z|D|2D|G|0|2|8|D|C|0|K|4|0|1|4|7|7|H|3|L|4|0|0|0|0|0|0|4|0|0|9|0|0|T|9|1A|4|1Q|3|2F|9|31|4|9|H|0|0|W|9|0|3|9|0^^$0|@$1|2|3|4|5|6|7|2D|8|@$9|2E|A|2F|B|C]|$9|2G|A|2H|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|2I|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|2J|8|@$9|2K|A|2L|B|C]|$9|2M|A|2N|B|C]|$9|2O|A|2P|B|C]]|D|@]|E|$]]|$1|M|3|N|5|H|7|2Q|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|2R|8|@]|D|@]|E|$]]|$1|Q|3|R|5|6|7|2S|8|@$9|2T|A|2U|B|C]|$9|2V|A|2W|B|C]]|D|@]|E|$]]|$1|S|3|T|5|6|7|2X|8|@$9|2Y|A|2Z|B|U]|$9|30|A|31|B|C]|$9|32|A|33|B|C]|$9|34|A|35|B|C]|$9|36|A|37|B|C]]|D|@]|E|$]]|$1|V|3|W|5|6|7|38|8|@]|D|@]|E|$]]|$1|X|3|Y|5|6|7|39|8|@]|D|@]|E|$]]|$1|Z|3|10|5|6|7|3A|8|@]|D|@]|E|$]]|$1|11|3|12|5|6|7|3B|8|@$9|3C|A|3D|B|C]|$9|3E|A|3F|B|C]|$9|3G|A|3H|B|C]]|D|@]|E|$]]|$1|13|3|14|5|6|7|3I|8|@$9|3J|A|3K|B|C]]|D|@]|E|$]]|$1|15|3|16|5|6|7|3L|8|@$9|3M|A|3N|B|C]]|D|@]|E|$]]|$1|17|3|18|5|6|7|3O|8|@$9|3P|A|3Q|B|C]]|D|@]|E|$]]|$1|19|3|1A|5|6|7|3R|8|@]|D|@]|E|$]]|$1|1B|3|1C|5|6|7|3S|8|@$9|3T|A|3U|B|C]|$9|3V|A|3W|B|C]|$9|3X|A|3Y|B|C]|$9|3Z|A|40|B|C]|$9|41|A|42|B|C]]|D|@]|E|$]]|$1|1D|3|1E|5|6|7|43|8|@$9|44|A|45|B|C]|$9|46|A|47|B|C]|$9|48|A|49|B|C]|$9|4A|A|4B|B|C]|$9|4C|A|4D|B|C]]|D|@]|E|$]]|$1|1F|3|1G|5|6|7|4E|8|@$9|4F|A|4G|B|C]|$9|4H|A|4I|B|C]]|D|@]|E|$]]|$1|1H|3|1I|5|6|7|4J|8|@$9|4K|A|4L|B|C]]|D|@]|E|$]]|$1|1J|3|1K|5|1L|7|4M|8|@$9|4N|A|4O|B|C]|$9|4P|A|4Q|B|C]|$9|4R|A|4S|B|C]|$9|4T|A|4U|B|C]]|D|@]|E|$]]|$1|1M|3|-4|5|6|7|4V|8|@]|D|@]|E|$]]|$1|1N|3|1O|5|H|7|4W|8|@]|D|@]|E|$I|J]]|$1|1P|3|1Q|5|6|7|4X|8|@]|D|@]|E|$]]|$1|1R|3|-4|5|6|7|4Y|8|@]|D|@]|E|$]]|$1|1S|3|1T|5|1U|7|4Z|8|@$9|50|A|51|B|C]]|D|@]|E|$]]|$1|1V|3|1W|5|1U|7|52|8|@$9|53|A|54|B|C]]|D|@]|E|$]]|$1|1X|3|1Y|5|1U|7|55|8|@]|D|@]|E|$]]|$1|1Z|3|20|5|1U|7|56|8|@$9|57|A|58|B|C]|$9|59|A|5A|B|C]|$9|5B|A|5C|B|C]|$9|5D|A|5E|B|C]|$9|5F|A|5G|B|C]]|D|@$9|5H|A|5I|1|5J]]|E|$]]|$1|21|3|22|5|1U|7|5K|8|@$9|5L|A|5M|B|C]]|D|@]|E|$]]|$1|23|3|24|5|6|7|5N|8|@$9|5O|A|5P|B|C]]|D|@]|E|$]]|$1|25|3|-4|5|6|7|5Q|8|@]|D|@]|E|$]]]|26|$27|$5|28|29|2A|E|$2B|2C]]]]

<code>String.contains(String)</code> or <code>String.indexOf(String)</code> - suggested
<pre class="lang-java prettyprint-override"><code>&quot;abc&quot;.contains(&quot;Z&quot;); // false - correct
&quot;zzzz&quot;.contains(&quot;Z&quot;); // false - correct
&quot;Z&quot;.contains(&quot;Z&quot;); // true - correct
&quot;and&quot;.contains(&quot;&quot;); // true - correct
&quot;and&quot;.contains(&quot;&quot;); // false - correct
&quot;and&quot;.indexOf(&quot;&quot;); // 0 - correct
&quot;and&quot;.indexOf(&quot;&quot;); // -1 - correct
</code></pre>
<code>String.indexOf(int)</code> and carefully considered <code>String.indexOf(char)</code> with char to <code>int</code> widening
<pre class="lang-java prettyprint-override"><code>&quot;and&quot;.indexOf(&quot;&quot;.charAt(0)); // 0 though incorrect usage has correct output due to portion of correct data
&quot;and&quot;.indexOf(&quot;&quot;.charAt(0)); // 0 -- incorrect usage and ambiguous result
&quot;and&quot;.indexOf(&quot;&quot;.codePointAt(0)); // -1 -- correct usage and correct output
</code></pre>
<h2>The discussions around character is ambiguous in Java world</h2>
<h3>can the value of <code>char</code> or <code>Character</code> considered as single character?</h3>
No. In the context of unicode characters, <code>char</code> or <code>Character</code> can sometimes be <code>part of a single character</code> and should not be treated as <code>a complete single character</code> logically.
<h3>if not, what should be considered as single character (logically)?</h3>
Any system supporting character encodings for Unicode characters should consider unicode's codepoint as single character.
So Java should do that very clear &amp; loud rather than exposing too much of internal implementation details to users.
<code>String</code> class is bad at abstraction (though it requires confusingly good amount of <code>understanding of its encapsulations to understand the abstraction </code> and hence an <code>anti-pattern</code>).
<h3>How is it different from general <code>char</code> usage?</h3>
<code>char</code> can be only be mapped to a character in Basic Multilingual Plane.
Only <code>codePoint - int</code> can cover the complete range of Unicode characters.
<h3>Why is this difference?</h3>
<code>char</code> is internally treated as <code>16-bit</code> unsigned value and could not represent all the unicode characters using UTF-16 internal representation using only <code>2-bytes</code>. Sometimes, values in a <code>16-bit</code> range have to be combined with another <code>16-bit</code> value to correctly define character.
Without getting too verbose, the usage of <code>indexOf</code>, <code>charAt</code>, <code>length</code> and such methods should be more explicit. Sincerely hoping Java will add new <code>UnicodeString</code> and <code>UnicodeCharacter</code> classes with clearly defined abstractions.
<h3>Reason to prefer <code>contains</code> and not <code>indexOf(int)</code></h3>
<ol>
<li>Practically there are many code flows that treat a logical character as <code>char</code> in java.</li>
<li>In Unicode context, <code>char</code> is not sufficient</li>
<li>Though the <code>indexOf</code> takes in an <code>int</code>, <code>char</code> to <code>int</code> conversion masks this from the user and user might do something like <code>str.indexOf(someotherstr.charAt(0))</code>(unless the user is aware of the exact context)</li>
<li>So, treating everything as <code>CharSequence</code> (aka <code>String</code>) is better</li>
</ol>
<pre class="lang-java prettyprint-override"><code> public static void main(String[] args) {
 System.out.println(&quot;and&quot;.indexOf(&quot;&quot;.charAt(0))); // 0 though incorrect usage has correct output due to portion of correct data
 System.out.println(&quot;and&quot;.indexOf(&quot;&quot;.charAt(0))); // 0 -- incorrect usage and ambiguous result
 System.out.println(&quot;and&quot;.indexOf(&quot;&quot;.codePointAt(0))); // -1 -- correct usage and correct output
 System.out.println(&quot;and&quot;.contains(&quot;&quot;)); // true - correct
 System.out.println(&quot;and&quot;.contains(&quot;&quot;)); // false - correct
 }
</code></pre>
<h2>Semantics</h2>
<ol>
<li><code>char</code> can handle most of the practical use cases. Still its better to use codepoints within programming environment for future extensibility.</li>
<li><code>codepoint</code> should handle nearly all of the technical use cases around encodings.</li>
<li>Still, <a href="https://unicode.org/reports/tr29/#Grapheme_Cluster_Boundaries" rel="nofollow noreferrer">Grapheme Clusters</a> falls out of the scope of <code>codepoint</code> level of abstraction.</li>
<li>Storage layers can choose <code>char</code> interface if <code>int</code>s are too costly(doubled). Unless storage cost is the only metric, its still better to use <code>codepoint</code>. Also, its better to treat storage as <code>byte</code> and delegate semantics to business logic built around storage.</li>
<li>Semantics can be abstracted at multiple levels. <code>codepoint</code> should become lowest level of interface and other semantics can be built around <code>codepoint</code> in runtime environment.</li>
</ol>

In Java is there a way to check the condition:

"Does this single character appear at all in string x"

without using a loop?

How can I check if a single character appears in a string?

数据结构

Java

在Java中，有一种检查条件的方法：“此单个字符是否出现在字符串x中”没有使用循环的的？

问如何检查字符串中是否出现单个字符？
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