blocks|key|1578486|text|每当您查找下一个匹配项时，递增lastIndex。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1578487|否则，它总是查找第一个子字符串(位置0)。|1578488|entityMap^0|F|9|0|0^^$0|@$1|2|3|4|5|6|7|J|8|@$9|K|A|L|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|M|8|@]|D|@]|E|$]]|$1|H|3|-4|5|6|7|N|8|@]|D|@]|E|$]]]|I|$]]

Increment <code>lastIndex</code> whenever you look for next occurrence. 

Otherwise it's always finding the first substring (at position 0).

blocks|key|1363347|text|尝试将lastIndex%2B=findStr.length()添加到循环的末尾，否则将陷入无休止的循环，因为一旦找到子字符串，您就会一次又一次地从相同的最后一个位置查找它。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1363348|entityMap^0|3|R|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|K|8|@]|D|@]|E|$]]]|G|$]]

try adding <code>lastIndex+=findStr.length()</code> to the end of your loop, otherwise you will end up in an endless loop because once you found the substring, you are trying to find it again and again from the same last position.

blocks|key|1578571|text|String+str+=+"helloslkhellodjladfjhello";
String+findStr+=+"hello";
int+lastIndex+=+0;
int+count+=+0;

while((lastIndex+=+str.indexOf(findStr,+lastIndex))+!=+-1)+{
+++++count%2B%2B;
+++++lastIndex+%2B=+findStr.length()+-+1;
}
System.out.println(count);|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1578572|循环结束时计数为3；希望这会有所帮助|unstyled|1578573|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|J|8|@]|9|@]|A|$]]|$1|G|3|-4|5|F|7|K|8|@]|9|@]|A|$]]]|H|$]]

<pre><code>String str = "helloslkhellodjladfjhello";
String findStr = "hello";
int lastIndex = 0;
int count = 0;

while((lastIndex = str.indexOf(findStr, lastIndex)) != -1) {
 count++;
 lastIndex += findStr.length() - 1;
}
System.out.println(count);
</code></pre>

at the end of the loop count is 3; hope it helps

blocks|key|1578601|text|你真的需要自己处理匹配吗？特别是如果你所需要的只是出现的次数，正则表达式会更整洁：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1578602|String+str+=+"helloslkhellodjladfjhello";
Pattern+p+=+Pattern.compile("hello");
Matcher+m+=+p.matcher(str);
int+count+=+0;
while+(m.find()){
++++count+%2B=1;
}
System.out.println(count);+++++|code-block|syntax|javascript|1578603|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|$]]

Do you really have to handle the matching yourself ? Especially if all you need is the number of occurences, regular expressions are tidier :

<pre><code>String str = "helloslkhellodjladfjhello";
Pattern p = Pattern.compile("hello");
Matcher m = p.matcher(str);
int count = 0;
while (m.find()){
 count +=1;
}
System.out.println(count); 
</code></pre>

blocks|key|1578633|text|您的lastIndex+%2B=+findStr.length();被放在括号之外，导致无限循环(当没有找到任何匹配项时，lastIndex始终为findStr.length())。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1578634|修复后的版本如下：|1578635|String+str+=+"helloslkhellodjladfjhello";
String+findStr+=+"hello";
int+lastIndex+=+0;
int+count+=+0;

while+(lastIndex+!=+-1)+{

++++lastIndex+=+str.indexOf(findStr,+lastIndex);

++++if+(lastIndex+!=+-1)+{
++++++++count%2B%2B;
++++++++lastIndex+%2B=+findStr.length();
++++}
}
System.out.println(count);|code-block|syntax|javascript|1578636|entityMap^0|2|U|1Z|G|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]|$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|T|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|U|8|@]|D|@]|E|$K|L]]|$1|M|3|-4|5|6|7|V|8|@]|D|@]|E|$]]]|N|$]]

Your <code>lastIndex += findStr.length();</code> was placed outside the brackets, causing an infinite loop (when no occurence was found, lastIndex was always to <code>findStr.length()</code>). 

Here is the fixed version :

<pre><code>String str = "helloslkhellodjladfjhello";
String findStr = "hello";
int lastIndex = 0;
int count = 0;

while (lastIndex != -1) {

 lastIndex = str.indexOf(findStr, lastIndex);

 if (lastIndex != -1) {
 count++;
 lastIndex += findStr.length();
 }
}
System.out.println(count);
</code></pre>

blocks|key|1578712|text|最后一行是在制造一个问题。lastIndex永远不会是-1，所以会有一个无限循环。这可以通过将最后一行代码移到if块中来修复。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1578713|String+str+=+"helloslkhellodjladfjhello";
String+findStr+=+"hello";
int+lastIndex+=+0;
int+count+=+0;

while(lastIndex+!=+-1){

++++lastIndex+=+str.indexOf(findStr,lastIndex);

++++if(lastIndex+!=+-1){
++++++++count+%2B%2B;
++++++++lastIndex+%2B=+findStr.length();
++++}
}
System.out.println(count);|code-block|syntax|javascript|1578714|entityMap^0|D|9|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

The last line was creating a problem. <code>lastIndex</code> would never be at -1, so there would be an infinite loop. This can be fixed by moving the last line of code into the if block.

<pre><code>String str = "helloslkhellodjladfjhello";
String findStr = "hello";
int lastIndex = 0;
int count = 0;

while(lastIndex != -1){

 lastIndex = str.indexOf(findStr,lastIndex);

 if(lastIndex != -1){
 count ++;
 lastIndex += findStr.length();
 }
}
System.out.println(count);
</code></pre>

blocks|key|1363565|text|一个更短的版本。;)|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1363566|String+str+=+"helloslkhellodjladfjhello";
String+findStr+=+"hello";
System.out.println(str.split(findStr,+-1).length-1);|code-block|syntax|javascript|1363567|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|$]]

A shorter version. ;)

<pre><code>String str = "helloslkhellodjladfjhello";
String findStr = "hello";
System.out.println(str.split(findStr, -1).length-1);
</code></pre>

blocks|key|1578816|text|使用Apache+Commons+Lang中的StringUtils.countMatches怎么样？|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1578817|String+str+=+"helloslkhellodjladfjhello";
String+findStr+=+"hello";

System.out.println(StringUtils.countMatches(str,+findStr));|code-block|syntax|javascript|1578818|这将输出以下内容：|1578819|3|1578820|entityMap|0|LINK|mutability|MUTABLE|url|http://commons.apache.org/proper/commons-lang/apidocs/org/apache/commons/lang3/StringUtils.html#countMatches(java.lang.CharSequence,%2520java.lang.CharSequence)^0|N|O|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@$A|V|B|W|1|X]]|C|$]]|$1|D|3|E|5|F|7|Y|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|6|7|Z|8|@]|9|@]|C|$]]|$1|K|3|L|5|F|7|10|8|@]|9|@]|C|$G|H]]|$1|M|3|-4|5|6|7|11|8|@]|9|@]|C|$]]]|N|$O|$5|P|Q|R|C|$S|T]]]]

How about using <a href="http://commons.apache.org/proper/commons-lang/apidocs/org/apache/commons/lang3/StringUtils.html#countMatches(java.lang.CharSequence,%20java.lang.CharSequence)" rel="noreferrer">StringUtils.countMatches</a> from Apache Commons Lang?

<pre><code>String str = "helloslkhellodjladfjhello";
String findStr = "hello";

System.out.println(StringUtils.countMatches(str, findStr));
</code></pre>

That outputs:

<pre><code>3
</code></pre>

blocks|key|1578864|text|试试这个吧。它用一个-替换所有的匹配项。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1578865|String+str+=+"helloslkhellodjladfjhello";
String+findStr+=+"hello";
int+numberOfMatches+=+0;
while+(str.contains(findStr)){
++++str+=+str.replaceFirst(findStr,+"-");
++++numberOfMatches%2B%2B;
}|code-block|syntax|javascript|1578866|如果你不想破坏你的str，你可以用同样的内容创建一个新的字符串：|1578867|String+str+=+"helloslkhellodjladfjhello";
String+strDestroy+=+str;
String+findStr+=+"hello";
int+numberOfMatches+=+0;
while+(strDestroy.contains(findStr)){
++++strDestroy+=+strDestroy.replaceFirst(findStr,+"-");
++++numberOfMatches%2B%2B;
}|1578868|执行此代码块后，这些将是您的值：|1578869|str+=+"helloslkhellodjladfjhello"
strDestroy+=+"-slk-djladfj-"
findStr+=+"hello"
numberOfMatches+=+3|1578870|entityMap^0|A|1|0|0|9|3|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@$9|V|A|W|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|X|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|Y|8|@$9|Z|A|10|B|C]]|D|@]|E|$]]|$1|M|3|N|5|H|7|11|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|12|8|@]|D|@]|E|$]]|$1|Q|3|R|5|H|7|13|8|@]|D|@]|E|$I|J]]|$1|S|3|-4|5|6|7|14|8|@]|D|@]|E|$]]]|T|$]]

Try this one. It replaces all the matches with a <code>-</code>.

<pre><code>String str = "helloslkhellodjladfjhello";
String findStr = "hello";
int numberOfMatches = 0;
while (str.contains(findStr)){
 str = str.replaceFirst(findStr, "-");
 numberOfMatches++;
}
</code></pre>

And if you don't want to destroy your <code>str</code> you can create a new string with the same content:

<pre><code>String str = "helloslkhellodjladfjhello";
String strDestroy = str;
String findStr = "hello";
int numberOfMatches = 0;
while (strDestroy.contains(findStr)){
 strDestroy = strDestroy.replaceFirst(findStr, "-");
 numberOfMatches++;
}
</code></pre>

After executing this block these will be your values:

<pre><code>str = "helloslkhellodjladfjhello"
strDestroy = "-slk-djladfj-"
findStr = "hello"
numberOfMatches = 3
</code></pre>

blocks|key|1578925|text|这是一个高级版本，用于计算令牌在用户输入的字符串中出现的次数：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1578926|public+class+StringIndexOf+{

++++public+static+void+main(String[]+args)+{

++++++++Scanner+scanner+=+new+Scanner(System.in);

++++++++System.out.println("Enter+a+sentence+please:+\n");
++++++++String+string+=+scanner.nextLine();

++++++++int+atIndex+=+0;
++++++++int+count+=+0;

++++++++while+(atIndex+!=+-1)
++++++++{
++++++++++++atIndex+=+string.indexOf("hello",+atIndex);

++++++++++++if(atIndex+!=+-1)
++++++++++++{
++++++++++++++++count%2B%2B;
++++++++++++++++atIndex+%2B=+5;
++++++++++++}
++++++++}

++++++++System.out.println(count);
++++}

}|code-block|syntax|javascript|1578927|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|$]]

Here is the advanced version for counting how many times the token occurred in a user entered string:

<pre><code>public class StringIndexOf {

 public static void main(String[] args) {

 Scanner scanner = new Scanner(System.in);

 System.out.println("Enter a sentence please: \n");
 String string = scanner.nextLine();

 int atIndex = 0;
 int count = 0;

 while (atIndex != -1)
 {
 atIndex = string.indexOf("hello", atIndex);

 if(atIndex != -1)
 {
 count++;
 atIndex += 5;
 }
 }

 System.out.println(count);
 }

}
</code></pre>

blocks|key|1578961|text|下面的方法显示了子字符串在整个字符串上重复的次数。希望用足给你：-|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1578962|++++String+searchPattern="aaa";+//+search+string
++++String+str="aaaaaababaaaaaa";+//+whole+string
++++int+searchLength+=+searchPattern.length();+
++++int+totalLength+=+str.length();+
++++int+k+=+0;
++++for+(int+i+=+0;+i+<+totalLength+-+searchLength+%2B+1;+i%2B%2B)+{
++++++++String+subStr+=+str.substring(i,+searchLength+%2B+i);
++++++++if+(subStr.equals(searchPattern))+{
+++++++++++k%2B%2B;
++++++++}

++++}|code-block|syntax|javascript|1578963|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|$]]

This below method show how many time substring repeat on ur whole string. Hope use full to you:-

<pre><code> String searchPattern="aaa"; // search string
 String str="aaaaaababaaaaaa"; // whole string
 int searchLength = searchPattern.length(); 
 int totalLength = str.length(); 
 int k = 0;
 for (int i = 0; i &lt; totalLength - searchLength + 1; i++) {
 String subStr = str.substring(i, searchLength + i);
 if (subStr.equals(searchPattern)) {
 k++;
 }

 }
</code></pre>

blocks|key|1363830|text|给出的正确答案对于计算像行返回这样的东西是不好的，而且过于冗长。以后的答案会更好，但所有的答案都可以简单地用|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1363831|str.split(findStr).length|code-block|syntax|javascript|1363832|它不会使用问题中的示例删除尾随匹配。|1363833|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

The answer given as correct is no good for counting things like line returns and is far too verbose. Later answers are better but all can be achieved simply with

<pre><code>str.split(findStr).length
</code></pre>

It does not drop trailing matches using the example in the question.

blocks|key|1363883|text|这里是另一种解决方案，不使用正则表达式/模式/匹配器，甚至不使用StringUtils。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1363884|String+str+=+"helloslkhellodjladfjhelloarunkumarhelloasdhelloaruhelloasrhello";
++++++++String+findStr+=+"hello";
++++++++int+count+=0;
++++++++int+findStrLength+=+findStr.length();
++++++++for(int+i=0;i<str.length();i%2B%2B){
++++++++++++if(findStr.startsWith(Character.toString(str.charAt(i)))){
++++++++++++++++if(str.substring(i).length()+>=+findStrLength){
++++++++++++++++++++if(str.substring(i,+i%2BfindStrLength).equals(findStr)){
++++++++++++++++++++++++count%2B%2B;
++++++++++++++++++++}
++++++++++++++++}
++++++++++++}
++++++++}
++++++++System.out.println(count);|code-block|syntax|javascript|1363885|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|$]]

here is the other solution without using regexp/patterns/matchers or even not using StringUtils.

<pre><code>String str = "helloslkhellodjladfjhelloarunkumarhelloasdhelloaruhelloasrhello";
 String findStr = "hello";
 int count =0;
 int findStrLength = findStr.length();
 for(int i=0;i&lt;str.length();i++){
 if(findStr.startsWith(Character.toString(str.charAt(i)))){
 if(str.substring(i).length() &gt;= findStrLength){
 if(str.substring(i, i+findStrLength).equals(findStr)){
 count++;
 }
 }
 }
 }
 System.out.println(count);
</code></pre>

blocks|key|1363916|text|如果您需要原始字符串中每个子字符串的索引，您可以对indexOf执行如下操作：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1363917|+private+static+List<Integer>+getAllIndexesOfSubstringInString(String+fullString,+String+substring)+{
++++int+pointIndex+=+0;
++++List<Integer>+allOccurences+=+new+ArrayList<Integer>();
++++while(fullPdfText.indexOf(substring,pointIndex)+>=+0){
+++++++allOccurences.add(fullPdfText.indexOf(substring,+pointIndex));
+++++++pointIndex+=+fullPdfText.indexOf(substring,+pointIndex)+%2B+substring.length();
++++}
++++return+allOccurences;
}|code-block|syntax|javascript|1363918|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 the index of each substring within the original string, you can do something with indexOf like this:

<pre><code> private static List&lt;Integer&gt; getAllIndexesOfSubstringInString(String fullString, String substring) {
 int pointIndex = 0;
 List&lt;Integer&gt; allOccurences = new ArrayList&lt;Integer&gt;();
 while(fullPdfText.indexOf(substring,pointIndex) &gt;= 0){
 allOccurences.add(fullPdfText.indexOf(substring, pointIndex));
 pointIndex = fullPdfText.indexOf(substring, pointIndex) + substring.length();
 }
 return allOccurences;
}
</code></pre>

blocks|key|1579088|text|正如@Mr_and_Mrs_D所建议的：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1579089|String+haystack+=+"hellolovelyworld";
String+needle+=+"lo";
return+haystack.split(Pattern.quote(needle),+-1).length+-+1;|code-block|syntax|javascript|1579090|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|$]]

As @Mr_and_Mrs_D suggested:

<pre><code>String haystack = "hellolovelyworld";
String needle = "lo";
return haystack.split(Pattern.quote(needle), -1).length - 1;
</code></pre>

blocks|key|1579108|text|下面是一个很好的、可重用的方法：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1579109|public+static+int+count(String+text,+String+find)+{
++++++++int+index+=+0;
++++++++int+count+=+0;
++++++++int+length+=+find.length();

++++++++while(+(index+=+text.indexOf(find,+index))+!=+-1+)+{++++++++++++++++
++++++++++++++++index+%2B=+length;+
++++++++++++++++count%2B%2B;
++++++++}

++++++++return+count;
}|code-block|syntax|javascript|1579110|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|$]]

Here it is, wrapped up in a nice and reusable method:
<pre><code>public static int count(String text, String find) {
 int index = 0;
 int count = 0;
 int length = find.length();

 while( (index = text.indexOf(find, index)) != -1 ) { 
 index += length; 
 count++;
 }

 return count;
}
</code></pre>

blocks|key|1579142|text|基于现有的答案，我想添加一个不带if的“简短”版本：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1579143|String+str+=+"helloslkhellodjladfjhello";
String+findStr+=+"hello";

int+count+=+0,+lastIndex+=+0;
while((lastIndex+=+str.indexOf(findStr,+lastIndex))+!=+-1)+{
++++lastIndex+%2B=+findStr.length()+-+1;
++++count%2B%2B;
}

System.out.println(count);+//+output:+3|code-block|syntax|javascript|1579144|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|$]]

Based on the existing answer(s) I'd like to add a "shorter" version without the if:

<pre><code>String str = "helloslkhellodjladfjhello";
String findStr = "hello";

int count = 0, lastIndex = 0;
while((lastIndex = str.indexOf(findStr, lastIndex)) != -1) {
 lastIndex += findStr.length() - 1;
 count++;
}

System.out.println(count); // output: 3
</code></pre>

blocks|key|1364072|text|您可以使用内置库函数执行以下操作：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1364073|import+org.springframework.util.StringUtils;
StringUtils.countOccurrencesOf(result,+"R-")|code-block|syntax|javascript|1364074|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|$]]

You can number of occurrences using inbuilt library function:

<pre><code>import org.springframework.util.StringUtils;
StringUtils.countOccurrencesOf(result, "R-")
</code></pre>

blocks|key|1364129|text|我很惊讶没有人提到这一点。它简单、简洁，性能略优于str.split(target,+-1).length-1|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1364130|public+static+int+count(String+str,+String+target)+{
++++return+(str.length()+-+str.replace(target,+"").length())+/+target.length();
}|code-block|syntax|javascript|1364131|entityMap^0|P|U|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

I'm very surprised no one has mentioned this one liner. It's simple, concise and performs slightly better than <code>str.split(target, -1).length-1</code>

<pre><code>public static int count(String str, String target) {
 return (str.length() - str.replace(target, "").length()) / target.length();
}
</code></pre>

blocks|key|1579255|text|public+int+countOfOccurrences(String+str,+String+subStr)+{
++return+(str.length()+-+str.replaceAll(Pattern.quote(subStr),+"").length())+/+subStr.length();
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1579256|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>public int countOfOccurrences(String str, String subStr) {
 return (str.length() - str.replaceAll(Pattern.quote(subStr), "").length()) / subStr.length();
}
</code></pre>

blocks|key|1364153|text|public+static+int+getCountSubString(String+str+,+String+sub){
int+n+=+0,+m+=+0,+counter+=+0,+counterSub+=+0;
while(n+<+str.length()){
++counter+=+0;
++m+=+0;
++while(m+<+sub.length()+&&+str.charAt(n)+==+sub.charAt(m)){
++++counter%2B%2B;
++++m%2B%2B;+n%2B%2B;
++}
++if+(counter+==+sub.length()){
++++counterSub%2B%2B;
++++continue;
++}
++else+if(counter+>+0){
++++continue;
++}
++n%2B%2B;
}

return++counterSub;|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1364154|}|unstyled|1364155|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|J|8|@]|9|@]|A|$]]|$1|G|3|-4|5|F|7|K|8|@]|9|@]|A|$]]]|H|$]]

<pre><code>public static int getCountSubString(String str , String sub){
int n = 0, m = 0, counter = 0, counterSub = 0;
while(n &lt; str.length()){
 counter = 0;
 m = 0;
 while(m &lt; sub.length() &amp;&amp; str.charAt(n) == sub.charAt(m)){
 counter++;
 m++; n++;
 }
 if (counter == sub.length()){
 counterSub++;
 continue;
 }
 else if(counter &gt; 0){
 continue;
 }
 n++;
}

return counterSub;
</code></pre>

}

blocks|key|1364192|text|此解决方案打印给定子字符串在整个字符串中出现的总次数，还包括存在重叠匹配的情况。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1364193|class+SubstringMatch{
++++public+static+void+main(String+[]args){
++++++++//String+str+=+"aaaaabaabdcaa";
++++++++//String+sub+=+"aa";
++++++++//String+str+=+"caaab";
++++++++//String+sub+=+"aa";
++++++++String+str="abababababaabb";
++++++++String+sub+=+"bab";

++++++++int+n+=+str.length();
++++++++int+m+=+sub.length();

++++++++//+index=-1+in+case+of+no+match,+otherwise+>=0(first+match+position)
++++++++int+index=str.indexOf(sub),+i=index%2B1,+count=(index>=0)?1:0;
++++++++System.out.println(i%2B"+"%2Bindex%2B"+"%2Bcount);

++++++++//+i+will+traverse+up+to+only+(m-n)+position
++++++++while(index!=-1+&&+i<=(n-m)){+++
++++++++++++index=str.substring(i,+n).indexOf(sub);
++++++++++++count=(index>=0)?count%2B1:count;
++++++++++++i=i%2Bindex%2B1;++
++++++++++++System.out.println(i%2B"+"%2Bindex);
++++++++}
++++++++System.out.println("count:+"%2Bcount);
++++}
}|code-block|syntax|javascript|1364194|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|$]]

This solution prints the total number of occurrence of a given substring throughout the string, also includes the cases where overlapping matches do exist.

<pre><code>class SubstringMatch{
 public static void main(String []args){
 //String str = "aaaaabaabdcaa";
 //String sub = "aa";
 //String str = "caaab";
 //String sub = "aa";
 String str="abababababaabb";
 String sub = "bab";

 int n = str.length();
 int m = sub.length();

 // index=-1 in case of no match, otherwise &gt;=0(first match position)
 int index=str.indexOf(sub), i=index+1, count=(index&gt;=0)?1:0;
 System.out.println(i+" "+index+" "+count);

 // i will traverse up to only (m-n) position
 while(index!=-1 &amp;&amp; i&lt;=(n-m)){ 
 index=str.substring(i, n).indexOf(sub);
 count=(index&gt;=0)?count+1:count;
 i=i+index+1; 
 System.out.println(i+" "+index);
 }
 System.out.println("count: "+count);
 }
}
</code></pre>

blocks|key|1364208|text|只是一个更好的答案|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1364209|++++public+int+countOccurrences(String+str,+String+sub)+{
++++++++if+(str+==+null+%7C%7C+str.length()+==+0+%7C%7C+sub+==+null+%7C%7C+sub.length()+==+0)+return+0;
++++++++int+count+=+0;
++++++++int+i+=+0;
++++++++while+((i+=+str.indexOf(sub,+i))+!=+-1)+{
++++++++++++count%2B%2B;
++++++++++++i+%2B=+sub.length();
++++++++}
++++++++return+count;
++++}|code-block|syntax|javascript|1364210|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|$]]

Just a little more peachy answer
<pre><code> public int countOccurrences(String str, String sub) {
 if (str == null || str.length() == 0 || sub == null || sub.length() == 0) return 0;
 int count = 0;
 int i = 0;
 while ((i = str.indexOf(sub, i)) != -1) {
 count++;
 i += sub.length();
 }
 return count;
 }
</code></pre>

blocks|key|1364221|text|我刚才在面试中被问到这个问题，我完全是一片空白。(像往常一样，我告诉自己，面试结束的那一刻我就会得到解决方案)，在电话结束5分钟后，我做到了:(|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1364222|++++int+subCounter=0;
++++int+count+=0;
++++for(int+i=0;+i<str.length();+i%2B%2B)+{
++++++++if((subCounter==0+&&+"a".equals(str.substring(i,i%2B1)))+
++++++++++++++++%7C%7C+(subCounter==1+&&+"b".equals(str.substring(i,i%2B1)))
++++++++++++++++%7C%7C+(subCounter==2+&&+"c".equals(str.substring(i,i%2B1))))+{
++++++++++++%2B%2BsubCounter;
++++++++}
++++++++if(subCounter==3)+{
++++++++++++count+=+count%2B1;
++++++++++++subCounter=0;
++++++++}
++++}
++++System.out.println(count);|code-block|syntax|javascript|1364223|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|$]]

I was asked this question in an interview just now and I went completely blank. (Like always, I told myself that the moment the interview ends ill get the solution) which I did, 5 mins after the call ended :(
<pre><code> int subCounter=0;
 int count =0;
 for(int i=0; i&lt;str.length(); i++) {
 if((subCounter==0 &amp;&amp; &quot;a&quot;.equals(str.substring(i,i+1))) 
 || (subCounter==1 &amp;&amp; &quot;b&quot;.equals(str.substring(i,i+1)))
 || (subCounter==2 &amp;&amp; &quot;c&quot;.equals(str.substring(i,i+1)))) {
 ++subCounter;
 }
 if(subCounter==3) {
 count = count+1;
 subCounter=0;
 }
 }
 System.out.println(count);
</code></pre>

Why is the following algorithm not halting for me?
(str is the string I am searching in, findStr is the string I am trying to find)

<pre><code>String str = "helloslkhellodjladfjhello";
String findStr = "hello";
int lastIndex = 0;
int count = 0;

while (lastIndex != -1) {
 lastIndex = str.indexOf(findStr,lastIndex);

 if( lastIndex != -1)
 count++;

 lastIndex += findStr.length();
}

System.out.println(count);
</code></pre>

Occurrences of substring in a string

为什么下面的算法对我来说没有停止？(str是我要搜索的字符串，findStr是我要查找的字符串)String str = "helloslkhellodjladfjhello";String findStr = "hello";int lastIndex = 0;int count = 0;while (lastInd...

问子字符串在字符串中的出现次数
EN

社区

活动

资源

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问子字符串在字符串中的出现次数EN

社区

活动

资源

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问子字符串在字符串中的出现次数
EN