blocks|key|4367334|text|在此语句中：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4367335|for+(String+s+:+paths){
++++++++++++s+=+nodevalue+%2B+s;
}|code-block|syntax|javascript|4367336|实际上，你并没有改变s的值，实际上，你不能这样做。for-each循环不能更改它正在迭代的元素。|4367337|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|$]]

In this statement:

<pre><code>for (String s : paths){
 s = nodevalue + s;
}
</code></pre>

You're not actually changing the value of s. In fact, you can't the way you're doing it now. A for-each loop can not change the elements that it is iterating over.

blocks|key|4367352|text|字符串是赋值的不可变类型|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4367353|s+=+nodevalue+%2B+s;|code-block|syntax|javascript|4367354|未被重新识别|4367355|更好的解决方案应该是|4367356|for+(TreeNode+t+:+childList){
++++final+List<String>+pathes+=+t.findStrings();
++++for+(final+String+path+:+pathes)+{
++++++//+add+all+pathes+to+paths+list+adding+nodevalue+to+the+beginning
++++++paths.add(nodevalue+%2B+path);+
++++}
}|4367357|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|$]]

String is an immutable type the assignment

<pre><code>s = nodevalue + s;
</code></pre>

is not regognized

A better solution should be 

<pre><code>for (TreeNode t : childList){
 final List&lt;String&gt; pathes = t.findStrings();
 for (final String path : pathes) {
 // add all pathes to paths list adding nodevalue to the beginning
 paths.add(nodevalue + path); 
 }
}
</code></pre>

blocks|key|893064|text|当您在内部循环中更改s时，您只是重新分配变量s，而不是存储在链表中的值。相反，您应该遍历列表中的所有元素，逐个更新它们。我认为像这样的东西应该是有效的：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|893065|//use+recursion+to+add+paths+from+all+children+to+the+list
for+(TreeNode+t+:+childList){
++++paths.addAll(t.findStrings());
++++//add+nodevalue+to+the+beginning+of+all+strings+in+the+list
++++int+length+=+paths.size();
++++for+(int+i=0;+i<length;+i%2B%2B)+{
++++++++paths.offer(nodevalue+%2B+paths.poll());
++++}
}|code-block|syntax|javascript|893066|poll获取列表前面的第一项，offer将结果放在后面。你取下第一个项目，改变它，然后把它放在背面-重复paths.size()次，你最终得到了按原始顺序更新的项目。|893067|entityMap^0|A|1|M|1|0|0|0|4|F|5|1G|C|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]|$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|T|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|U|8|@$9|V|A|W|B|C]|$9|X|A|Y|B|C]|$9|Z|A|10|B|C]]|D|@]|E|$]]|$1|M|3|-4|5|6|7|11|8|@]|D|@]|E|$]]]|N|$]]

When you change <code>s</code> in the inner loop, you're just re-assigning the variable <code>s</code>, not the value stored in your linked list. Instead, you should loop through all the elements in your list updating them one by one. I think something like this should work:

<pre><code>//use recursion to add paths from all children to the list
for (TreeNode t : childList){
 paths.addAll(t.findStrings());
 //add nodevalue to the beginning of all strings in the list
 int length = paths.size();
 for (int i=0; i&lt;length; i++) {
 paths.offer(nodevalue + paths.poll());
 }
}
</code></pre>

<code>poll</code> takes the first item from the front of the list, and <code>offer</code> puts the result on the back. You take the first item off, change it, then put it on the back—repeat <code>paths.size()</code> times and you end up with the updated items in the original order.

blocks|key|2900364|text|增强的for循环在这里帮不上忙。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2900365|你将不得不使用传统的方法，如下所示：|2900366|for+(int+i=0;+i<paths.size();+i%2B%2B){
++++paths.set(i,+paths.get(i)+%2B+paths.get(i));
}|code-block|syntax|javascript|2900367|这里：public+E+set(int+index,+E+element)|offset|length|style|CODE|2900368|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.oracle.com/javase/6/docs/api/java/util/LinkedList.html#set%2528int,%2520E%2529^0|0|0|0|3|Y|3|Y|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|X|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|Y|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|Z|8|@$K|10|L|11|M|N]]|9|@$K|12|L|13|1|14]]|A|$]]|$1|O|3|-4|5|6|7|15|8|@]|9|@]|A|$]]]|P|$Q|$5|R|S|T|A|$U|V]]]]

The enhanced for loop is not going to help here.

You are going to have to go with the traditional one as follows:

<pre><code>for (int i=0; i&lt;paths.size(); i++){
 paths.set(i, paths.get(i) + paths.get(i));
}
</code></pre>

Here: <a href="http://docs.oracle.com/javase/6/docs/api/java/util/LinkedList.html#set%28int,%20E%29" rel="nofollow"><code>public E set(int index, E element)</code></a>

blocks|key|584175|text|正如Michael所说，您不能在for+each循环中替换条目的值。因为字符串是不可变的，所以你不能改变现有的字符串。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|584176|你需要执行一个普通的循环：|584177|for+(int+i=0;i<paths.size();+i%2B%2B)+{
++++paths.set(i,+nodevalue+%2Bpaths.get(i));
}|code-block|syntax|javascript|584178|请注意，您不会更改现有字符串的值，而是将其替换为同一位置的新字符串。|584179|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|N|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|O|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|P|8|@]|9|@]|A|$]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

as Michael said you can't replace the values of an entry in a for each loop. And since Strings are immutable you can't change the existing string.

You'll need to perform a normal Loop:

<pre><code>for (int i=0;i&lt;paths.size(); i++) {
 paths.set(i, nodevalue +paths.get(i));
}
</code></pre>

Note that you don't change the value of the existing String but replace it with a new String at the same location.

I'm trying to create a method that returns a linked list of Strings. There is a tree, and each node in the tree stores a character. The method is supposed to find all possible paths through the tree. Each path creates a string, which is added to the list. 

There seems to be a problem within the second for loop that I can't figure out. The method only returns characters that were added in the first if statement.

each node contains the variables childList, which is a linked list of children nodes, and nodevalue, which is the character that the node is storing.

<pre><code>public LinkedList&lt;String&gt; findStrings() {
 LinkedList&lt;String&gt; paths = new LinkedList&lt;String&gt;();
 //add character to list if there are no children
 if (childList.isEmpty()){
 paths.add("" + nodevalue);
 return paths;
 }
 //use recursion to add paths from all children to the list
 for (TreeNode t : childList){
 paths.addAll(t.findStrings());
 //add nodevalue to the beginning of all strings in the list
 for (String s : paths){
 s = nodevalue + s;
 }
 }
 for (String s : paths) System.out.println(s); //for debugging
 return paths;
}
</code></pre>

Looping through a list to add characters to a list of Strings 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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我正在尝试创建一个返回字符串链接列表的方法。有一棵树，树中的每个节点都存储一个字符。该方法被认为是通过树找到所有可能的路径。每条路径创建一个字符串，并将其添加到列表中。在第二个for循环中似乎有一个我无法解决的问题。该方法仅返回在第一个if语句中添加的字符。每个节点都包含变量childList和nodevalue，前者...

问循环遍历列表以向Java中的字符串列表添加字符
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