blocks|key|2566979|text|一种方法是创建一个新列表，然后循环第一个列表，在新列表的开头添加偶数，在末尾添加奇数。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2566980|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|$]]

One way is to create a new list and then loop through your first list, adding even numbers to the beginning of the new list and odd numbers to the end.

blocks|key|3903520|text|一个简单的解决方案是枚举列表中的所有元素，并将它们分配给两个不同的列表，例如even_list和odd_list，具体取决于数字的奇异性。然后使用基本排序分别对每个列表进行排序，最后将两个列表连接成一个新列表。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3903521|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|$]]

A simple solution is to enumerate all elements of the list and assign them to two different list, say even_list and odd_list depending on the oddity of the numbers. Then sort each list individually using basic sort and finally concatenate the two lists into a new list.

blocks|key|654396|text|我只是把这个列表浏览了两遍：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|654397|654398|第一次通过输出evens|unordered-list-item|654399|第二次通过输出odds|654400|654401|这将是O(n)，而使用比较器等可能不是。|654402|entityMap^0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|N|8|@]|9|@]|A|$]]|$1|C|3|D|5|E|7|O|8|@]|9|@]|A|$]]|$1|F|3|G|5|E|7|P|8|@]|9|@]|A|$]]|$1|H|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|R|8|@]|9|@]|A|$]]|$1|K|3|-4|5|6|7|S|8|@]|9|@]|A|$]]]|L|$]]

I'd just run through the list twice: 

<ul>
<li>first time through output the evens</li>
<li>second time through output the odds</li>
</ul>

This is going to be O(n) which using comparators etc may not be.

blocks|key|123963|text|++++LinkedList<Integer>+list+=+new+LinkedList<Integer>();
++++LinkedList<Integer>+newlist+=+new+LinkedList<Integer>();
++++int[]+a+=+{1,2,3,5,6,11,10};
++++for(int+i=0;i<a.length;i%2B%2B)+{
++++++++list.add(a[i]);
++++}

++++for(int+i=0,j=0;+i<list.size();+i%2B%2B)+{
++++++++if(a[i]%252+==+0)+{
++++++++++++newlist.add(j%2B%2B,+a[i]);
++++++++}+else+{
++++++++++++newlist.addLast(a[i]);
++++++++}
++++}

++++System.out.println(newlist);|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|123964|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> LinkedList&lt;Integer&gt; list = new LinkedList&lt;Integer&gt;();
 LinkedList&lt;Integer&gt; newlist = new LinkedList&lt;Integer&gt;();
 int[] a = {1,2,3,5,6,11,10};
 for(int i=0;i&lt;a.length;i++) {
 list.add(a[i]);
 }

 for(int i=0,j=0; i&lt;list.size(); i++) {
 if(a[i]%2 == 0) {
 newlist.add(j++, a[i]);
 } else {
 newlist.addLast(a[i]);
 }
 }

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

blocks|key|654419|text|我认为为了更好的操作使用ArrayList，你可以参考下面的方法，它返回你的排序列表。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|654420|public+static+List<Integer>+sortList(List<Integer>+list){
++++LinkedList<Integer>+sortedList+=+new+LinkedList<Integer>();

++++List<Integer>+evenList=new+ArrayList<Integer>();
++++List<Integer>+oddList=new+ArrayList<Integer>();
++++for(Integer+i:list){

++++++++if(i%252==0)
++++++++++++evenList.add(i);
++++++++else
++++++++++++oddList.add(i);
++++}

++++Collections.sort(evenList);
++++Collections.sort(oddList);
++++sortedList.addAll(evenList);
++++sortedList.addAll(oddList);


++++return+sortedList;
}|code-block|syntax|javascript|654421|输入：|654422|[1,+2,+3,+5,+6,+11,+10]|654423|输出：|654424|[2,+6,+10,+1,+3,+5,+11]|654425|entityMap^0|C|9|0|0|0|0|N|0|0|0|N|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]]|$1|K|3|L|5|6|7|Y|8|@]|D|@]|E|$]]|$1|M|3|N|5|6|7|Z|8|@$9|10|A|11|B|C]]|D|@]|E|$]]|$1|O|3|P|5|6|7|12|8|@]|D|@]|E|$]]|$1|Q|3|R|5|6|7|13|8|@$9|14|A|15|B|C]]|D|@]|E|$]]|$1|S|3|-4|5|6|7|16|8|@]|D|@]|E|$]]]|T|$]]

I think for manipulation better to use <code>ArrayList</code> , you can refer below method which returns your sorted List.

<pre><code>public static List&lt;Integer&gt; sortList(List&lt;Integer&gt; list){
 LinkedList&lt;Integer&gt; sortedList = new LinkedList&lt;Integer&gt;();

 List&lt;Integer&gt; evenList=new ArrayList&lt;Integer&gt;();
 List&lt;Integer&gt; oddList=new ArrayList&lt;Integer&gt;();
 for(Integer i:list){

 if(i%2==0)
 evenList.add(i);
 else
 oddList.add(i);
 }

 Collections.sort(evenList);
 Collections.sort(oddList);
 sortedList.addAll(evenList);
 sortedList.addAll(oddList);


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

Input:

<code>[1, 2, 3, 5, 6, 11, 10]</code>

Output:

<code>[2, 6, 10, 1, 3, 5, 11]</code>

blocks|key|2574554|text|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2574555|在temp中获取头部指针，一旦遇到正数，就使用temp+|unordered-list-item|2574556|遍历列表，在头部创建一个新节点，复制数据并删除temp指向的节点。头将指向新节点，临时将指向下一个节点。|2574557|遍历列表直到temp->next+!=+NULL|2574558|时间复杂度将为O(n)，因为创建和删除操作的复杂度为O(1)|2574559|2574560|entityMap^0|0|0|0|0|0|0^^$0|@$1|2|3|-4|4|5|6|M|7|@]|8|@]|9|$]]|$1|A|3|B|4|C|6|N|7|@]|8|@]|9|$]]|$1|D|3|E|4|C|6|O|7|@]|8|@]|9|$]]|$1|F|3|G|4|C|6|P|7|@]|8|@]|9|$]]|$1|H|3|I|4|C|6|Q|7|@]|8|@]|9|$]]|$1|J|3|-4|4|5|6|R|7|@]|8|@]|9|$]]|$1|K|3|-4|4|5|6|S|7|@]|8|@]|9|$]]]|L|$]]

<ul>
<li>Take head pointer in temp and traverse the list using temp</li>
<li>As soon as you encounter positive number, create a new node at head, copy the data and delete the node pointed by temp. Head will be pointing to new node, Temp will be pointing to next node. </li>
<li>Traverse the list until temp->next != NULL</li>
<li>Time Complexity will be O(n) since creation and deletion operation are O(1)</li>
</ul>

blocks|key|2574562|text|下面是Algo/实现|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2574563|public+static+LinearNode<Integer>+seperateEvenAndOddNodes(LinearNode<Integer>+head)+{
++++LinearNode<Integer>+tail+=+head,+prevNode=null,+currNode+=+head,+nextNode;
++++int+length+=+length(head),+count+=+0;
++++boolean+allEven+=+true,+allOdd+=+true;
++++//point+to+the+last+node,+start+dumping+all+nodes+after+this
++++while+(tail.next()+!=+null)+{
++++++++if+(tail.getElement()+%25+2+==+0)+{
++++++++++++allOdd+=+false;
++++++++}+else+{
++++++++++++allEven+=+false;
++++++++}
++++++++tail+=+tail.next();
++++}
++++//+Dont+do+anything+if+either+all+odd+or+all+even
++++if+(allOdd+%7C%7C+allEven)+{
++++++++return+head;
++++}
++++//+Make+sure+you+don't+go+in+infinite+loop,+and+hence+condition+to+make+sure,+you+traverse+limited+nodes.
++++while+(currNode+!=+null+&&+count+<+length)+{
++++++++nextNode+=+currNode.next();
++++++++//move+currNode+to+the+end+of+list,+if+it+is+odd.
++++++++if+(currNode.getElement()+%25+2+==+1)+{
++++++++++++LinearNode<Integer>+temp+=+currNode;
++++++++++++if+(prevNode+!=+null)+{
++++++++++++++++prevNode.next(nextNode);
++++++++++++++++currNode+=+prevNode;
++++++++++++}+else+{
++++++++++++++++head+=+nextNode;
++++++++++++++++currNode+=+null;
++++++++++++}
++++++++++++tail.next(temp);
++++++++++++tail+=+temp;
++++++++++++temp.next(null);
++++++++}
++++++++prevNode+=+currNode;
++++++++currNode+=+nextNode;
++++++++count%2B%2B;
++++}
++++//return+the+new+head,+in+case+the+list+begins+with+odd
++++return+head;
}|code-block|syntax|javascript|2574564|以下是单元测试|2574565|@Test
public+void+seperateEvenAndOddNodesTest()+{
++++LinearNode<Integer>+head+=+buildLinkedList(2,3,4,1,7,8,9);
++++head+=+LinkedListUtil.seperateEvenAndOddNodes(head);
++++assertLinkedList(head,+2,4,8,3,1,7,9);

++++head+=+buildLinkedList(9,3,4,1,7,8,2);
++++head+=+LinkedListUtil.seperateEvenAndOddNodes(head);
++++assertLinkedList(head,+4,8,2,9,3,1,7);


++++head+=+buildLinkedList(1,3,5,7);
++++head+=+LinkedListUtil.seperateEvenAndOddNodes(head);
++++assertLinkedList(head,+1,3,5,7);

++++head+=+buildLinkedList(2,4,6,8);
++++head+=+LinkedListUtil.seperateEvenAndOddNodes(head);
++++assertLinkedList(head,+2,4,6,8);
}|2574566|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

Here is the Algo/Implementation

<pre><code>public static LinearNode&lt;Integer&gt; seperateEvenAndOddNodes(LinearNode&lt;Integer&gt; head) {
 LinearNode&lt;Integer&gt; tail = head, prevNode=null, currNode = head, nextNode;
 int length = length(head), count = 0;
 boolean allEven = true, allOdd = true;
 //point to the last node, start dumping all nodes after this
 while (tail.next() != null) {
 if (tail.getElement() % 2 == 0) {
 allOdd = false;
 } else {
 allEven = false;
 }
 tail = tail.next();
 }
 // Dont do anything if either all odd or all even
 if (allOdd || allEven) {
 return head;
 }
 // Make sure you don't go in infinite loop, and hence condition to make sure, you traverse limited nodes.
 while (currNode != null &amp;&amp; count &lt; length) {
 nextNode = currNode.next();
 //move currNode to the end of list, if it is odd.
 if (currNode.getElement() % 2 == 1) {
 LinearNode&lt;Integer&gt; temp = currNode;
 if (prevNode != null) {
 prevNode.next(nextNode);
 currNode = prevNode;
 } else {
 head = nextNode;
 currNode = null;
 }
 tail.next(temp);
 tail = temp;
 temp.next(null);
 }
 prevNode = currNode;
 currNode = nextNode;
 count++;
 }
 //return the new head, in case the list begins with odd
 return head;
}
</code></pre>

Here is the unit tests

<pre><code>@Test
public void seperateEvenAndOddNodesTest() {
 LinearNode&lt;Integer&gt; head = buildLinkedList(2,3,4,1,7,8,9);
 head = LinkedListUtil.seperateEvenAndOddNodes(head);
 assertLinkedList(head, 2,4,8,3,1,7,9);

 head = buildLinkedList(9,3,4,1,7,8,2);
 head = LinkedListUtil.seperateEvenAndOddNodes(head);
 assertLinkedList(head, 4,8,2,9,3,1,7);


 head = buildLinkedList(1,3,5,7);
 head = LinkedListUtil.seperateEvenAndOddNodes(head);
 assertLinkedList(head, 1,3,5,7);

 head = buildLinkedList(2,4,6,8);
 head = LinkedListUtil.seperateEvenAndOddNodes(head);
 assertLinkedList(head, 2,4,6,8);
}
</code></pre>

blocks|key|2574568|text|在这里，我发布了一个根据位置而不是值排列节点的逻辑。通过适当地放置指针节点，您可以对您的问题使用类似的逻辑。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2574569|public+ListNode+oddEvenList(ListNode+head)+{
//if+null+
+if(head+==+null)+return+null;

//only+one/two+node/s+present
+if(head.next+==+null+%7C%7C+head.next.next+==+null)+return+head;

ListNode+even+=+head,+odd+=+even.next,+res+=+odd;
while(odd!=null+&&+even.next+!=+null+&&+odd.next+!=+null){
++++even.next+=+odd.next;
++++even+=+even.next;
++++odd.next+=+even.next;
++++odd+=+odd.next;
++++even.next+=+res;
}
return+head;
}|code-block|syntax|javascript|2574570|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 I'm posting a logic of arranging the nodes based on positions rather than values. You can use similar logic for your problem by placing the pointer nodes appropriately.

<pre><code>public ListNode oddEvenList(ListNode head) {
//if null 
 if(head == null) return null;

//only one/two node/s present
 if(head.next == null || head.next.next == null) return head;

ListNode even = head, odd = even.next, res = odd;
while(odd!=null &amp;&amp; even.next != null &amp;&amp; odd.next != null){
 even.next = odd.next;
 even = even.next;
 odd.next = even.next;
 odd = odd.next;
 even.next = res;
}
return head;
}
</code></pre>

Given a linked list, say {1,2,3,5,6,11,10}
I need the output as {2,6,10,1,3,5,11}.
The even numbers need to be arranged before the odd numbers.

how to arrange all even numbers in front of odd numbers in a linked list

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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社区新版编辑器体验调研

给定一个链表，例如{1,2,3,5,6,11,10}，我需要输出为{2,6,10,1,3,5,11}。偶数需要排在奇数之前。

问如何在链表中排列奇数前面的所有偶数
EN

回答 8

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问如何在链表中排列奇数前面的所有偶数EN

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问如何在链表中排列奇数前面的所有偶数
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