blocks|key|696828|text|在循环外部定义int+j+=+0，然后将System.out.println(y[j]+%2B+":+"+%2B+count);移到第一个for循环外部，并在外部for循环的第一行中将count重置为0。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|696829|顺便说一句，你为什么不使用String#equals|696830|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.oracle.com/javase/7/docs/api/java/lang/String.html#equals%2528java.lang.Object%2529^0|7|9|K|14|1T|3|25|3|2G|5|0|D|D|0|0^^$0|@$1|2|3|4|5|6|7|P|8|@$9|Q|A|R|B|C]|$9|S|A|T|B|C]|$9|U|A|V|B|C]|$9|W|A|X|B|C]|$9|Y|A|Z|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|10|8|@]|D|@$9|11|A|12|1|13]]|E|$]]|$1|H|3|-4|5|6|7|14|8|@]|D|@]|E|$]]]|I|$J|$5|K|L|M|E|$N|O]]]]

Define <code>int j = 0</code> outside the loop, then move <code>System.out.println(y[j] + ": " + count);</code> to outside the first <code>for</code> loop and reset <code>count</code> to 0 in the first line of the outer <code>for</code> loop.

BTW, why don't you use <a href="http://docs.oracle.com/javase/7/docs/api/java/lang/String.html#equals%28java.lang.Object%29" rel="nofollow">String#equals</a>?

blocks|key|696832|text|public+static+void+stringOccurInY(String[]+x,+String[]+y)+{
++++++++int+count+=+0;
++++++++for(int+i=0;+i<x.length;+i%2B%2B)++{
++++++++++++for(int+j=0;+j<y.length;+j%2B%2B)+{
++++++++++++++++if(y[j].contains(x[i]))+{
++++++++++++++++++++count%2B%2B;
++++++++++++++++}
++++++++++++}++++
++++++++++++System.out.println(y[j]+%2B+":+"+%2B+count);
++++++++++++count+=+0;+//+reset+the+count
++++++++}
++++}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|696833|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 static void stringOccurInY(String[] x, String[] y) {
 int count = 0;
 for(int i=0; i&lt;x.length; i++) {
 for(int j=0; j&lt;y.length; j++) {
 if(y[j].contains(x[i])) {
 count++;
 }
 } 
 System.out.println(y[j] + ": " + count);
 count = 0; // reset the count
 }
 }
</code></pre>

blocks|key|696846|text|进行以下修改以使其正常工作：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|696847|696848|交换机x和y在for环路中。|unordered-list-item|offset|length|style|CODE|696849|在外部循环中打印，以避免重复打印一次。|696850|在外部循环中初始化count，以避免重复初始化它。|696851|696852|公共类stringOccurInArray+{+public+static+void+(+String[]+x，String[]+y)+{+for(int+i=0；i|696853|entityMap^0|0|0|3|1|5|1|7|3|0|0|9|5|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|T|8|@]|9|@]|A|$]]|$1|C|3|D|5|E|7|U|8|@$F|V|G|W|H|I]|$F|X|G|Y|H|I]|$F|Z|G|10|H|I]]|9|@]|A|$]]|$1|J|3|K|5|E|7|11|8|@]|9|@]|A|$]]|$1|L|3|M|5|E|7|12|8|@$F|13|G|14|H|I]]|9|@]|A|$]]|$1|N|3|-4|5|6|7|15|8|@]|9|@]|A|$]]|$1|O|3|P|5|6|7|16|8|@]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|17|8|@]|9|@]|A|$]]]|R|$]]

Make the following modifications to get it to work right:
<li>Switch <code>x</code> and <code>y</code> in the <code>for</code> loops.
<li>Print in the outer loop to avoid repeating prints for one occurrence.
<li>Initialize <code>count</code> inside the outer loop to avoid repeatedly initializing it.

<pre><code>public class stringOccurInArray {
 public static void stringOccurInY(String[] x, String[] y) {
 for(int i=0; i&lt;y.length; i++) {
 int count = 0;
 for(int j=0; j&lt;x.length; j++) {
 if(x[j].equals(y[i])) {
 count++;
 }
 }
 System.out.println(y[i] + ": " + count);
 }
 }

 public static void main(String[] args) {
 String[] a = {"ab", "cd", "ab", "dc", "cd"};
 String[] b = {"ab", "dc", "ef"};

 stringOccurInY(a, b);
 }
}
</code></pre>

blocks|key|2030470|text|你的代码应该是：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2030471|for(int+i=0;+i<x.length;+i%2B%2B)++{+
count+=+0;
for(int+j=0;+j<y.length;+j%2B%2B)+{|code-block|syntax|javascript|2030472|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|$]]

Your code should be :

<pre><code>for(int i=0; i&lt;x.length; i++) { 
count = 0;
for(int j=0; j&lt;y.length; j++) {
</code></pre>

blocks|key|4155281|text|有几件事要提一下。像这样重写你的代码会更容易：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4155282|public+static+void+stringOccurInY(String[]+x,+String[]+y)+{
++++int+count+=+0;
++++for+(int+i+=+0;+i+<+y.length;+i%2B%2B)+{++++++++++
++++++++for+(int+j+=+0;+j+<+x.length;+j%2B%2B)+{
++++++++++++if+(y[i].contains(x[j]))+{
++++++++++++++++count%2B%2B;
++++++++++++}
++++++++}
++++++++System.out.println(y[i]+%2B+":+"+%2B+count);
++++++++count+=+0;+//+reset+the+count
++++}
}|code-block|syntax|javascript|4155283|你应该首先迭代y。|4155284|你也可以通过foreach循环来代替迭代。|4155285|for+(String+aY+:+y)+{
++++int+count+=+0;
++++for+(String+aX+:+x)+{
++++++++if+(aY.contains(aX))+{
++++++++++++count%2B%2B;
++++++++}
++++}
++++System.out.println(aY+%2B+":+"+%2B+count);
++++//no+need+to+reset+the+count
}|4155286|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]]|$1|G|3|H|5|6|7|Q|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|R|8|@]|9|@]|A|$]]|$1|K|3|L|5|D|7|S|8|@]|9|@]|A|$E|F]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

There are several things to mention. It's easier to rewrite your code like this:

<pre><code>public static void stringOccurInY(String[] x, String[] y) {
 int count = 0;
 for (int i = 0; i &lt; y.length; i++) { 
 for (int j = 0; j &lt; x.length; j++) {
 if (y[i].contains(x[j])) {
 count++;
 }
 }
 System.out.println(y[i] + ": " + count);
 count = 0; // reset the count
 }
}
</code></pre>

you should iterate over y in the first place.

Also you could replace iterating via foreach loop.

<pre><code>for (String aY : y) {
 int count = 0;
 for (String aX : x) {
 if (aY.contains(aX)) {
 count++;
 }
 }
 System.out.println(aY + ": " + count);
 //no need to reset the count
}
</code></pre>

blocks|key|1838396|text|你也可以试试下面的代码。做你想做的事：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1838397|public+static+void+stringOccurInY(String[]+x,+String[]+y)+{
++++++++int+count+=+0;
++++++++HashMap<String,+Integer>+map+=+new+HashMap<String,+Integer>();
++++++++for(int+i=0;+i<x.length;+i%2B%2B)++{
++++++++++++for(int+j=0;+j<y.length;+j%2B%2B)+{
++++++++++++++++if(y[j].contains(x[i]))+{
+++++++++++++++++++++if(map.get(y[j])+!=+null){
+++++++++++++++++++++++++count+=+(int)+map.get(y[j]);
+++++++++++++++++++++++++count%2B%2B;
+++++++++++++++++++++++++map.put(y[j],+count);
+++++++++++++++++++++}else{
+++++++++++++++++++++++++map.put(y[j],+1);
+++++++++++++++++++++}

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

++++++++}
++++++++System.out.println(map);
++++}|code-block|syntax|javascript|1838398|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 try below code as well. Does exactly what you want:

<pre><code>public static void stringOccurInY(String[] x, String[] y) {
 int count = 0;
 HashMap&lt;String, Integer&gt; map = new HashMap&lt;String, Integer&gt;();
 for(int i=0; i&lt;x.length; i++) {
 for(int j=0; j&lt;y.length; j++) {
 if(y[j].contains(x[i])) {
 if(map.get(y[j]) != null){
 count = (int) map.get(y[j]);
 count++;
 map.put(y[j], count);
 }else{
 map.put(y[j], 1);
 }

 }
 }

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

I created a Java method which when passed two arrays of strings x and y, counts the number of times each string occurring in y occurs in x and prints the results in the order the strings appear in y. For example, look at the main function, it should output as ab: 2, dc: 1, ef: 0. My code didn't work as it outputs ab: 1, ab: 2, dc: 3.

<pre><code>public class stringOccurInArray {
 public static void stringOccurInY(String[] x, String[] y) {
 int count = 0;
 for(int i=0; i&lt;x.length; i++) {
 for(int j=0; j&lt;y.length; j++) {
 if(y[j].contains(x[i])) {
 count++;
 System.out.println(y[j] + ": " + count);
 }
 }
 }
 count = 0; // reset the count
 }

 public static void main(String[] args) {
 String[] a = {"ab", "cd", "ab", "dc", "cd"};
 String[] b = {"ab", "dc", "ef"};

 stringOccurInY(a, b);
 }
}
</code></pre>

Pass two arrays of strings and count the occur y array in x array in Java

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我创建了一个Java方法，当传递两个字符串x和y数组时，计算每个字符串在y中出现的次数，并按照字符串在y中出现的顺序打印结果。例如，看看main函数，它应该输出为ab: 2，dc: 1，ef: 0。我的代码不工作，因为它输出ab: 1，ab: 2，dc: 3。public class stringOccurInArra...

问在Java中传递两个字符串数组，并计算x数组中出现的y数组
EN

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问在Java中传递两个字符串数组，并计算x数组中出现的y数组EN