# LeetCode – Partition List (Java)

Given a linked list and a value x, partition it such that all nodes less than x come before nodes greater than or equal to x.

You should preserve the original relative order of the nodes in each of the two partitions.

For example, given 1->4->3->2->5->2 and x = 3, return 1->2->2->4->3->5.

Java Solution

```public class Solution { public ListNode partition(ListNode head, int x) { if(head == null) return null;   ListNode fakeHead1 = new ListNode(0); ListNode fakeHead2 = new ListNode(0); fakeHead1.next = head;   ListNode p = head; ListNode prev = fakeHead1; ListNode p2 = fakeHead2;   while(p != null){ if(p.val < x){ p = p.next; prev = prev.next; }else{   p2.next = p; prev.next = p.next;   p = prev.next; p2 = p2.next; } }   // close the list p2.next = null;   prev.next = fakeHead2.next;   return fakeHead1.next; } }```
Category >> Algorithms
If you want someone to read your code, please put the code inside <pre><code> and </code></pre> tags. For example:
```<pre><code>
String foo = "bar";
</code></pre>
```

The result is correct indeed

• Jeff Guo

This one might be easier to understand. C# code.

public Node PartitionList(Node head, T number)

{

var fakeHead1 = new Node(); //smaller ones

var fakeHead2 = new Node(); //bigger ones

while (p != null)

{

if (p.content.CompareTo(number) < 0)

{

fakeTail1.Next = p;

fakeTail1 = fakeTail1.Next;

p = p.Next;

fakeTail1.Next = null;

}

else

{

faketail2.Next = p;

faketail2 = faketail2.Next;

p = p.Next;

faketail2.Next = null;

}

}

}

• Akhil Verghese

Good answer, but if you’re going to swap the p and q values, the q value should be < num. You're skipping passed all the values num), it’ll work (tested and confirmed).

• Antoine Dahan

Here is a much simpler solution:

I’m iterating through with two iterators so I can change the current positions previous node’s next (no other way to access it in a singly linked list). Then whenever I encounter a value less than x, I append it in front of the head and set it as the new head. If the value is greater than or equal to x I just go onto the next value.

After reaching the end, I just return the new head.

(posted as screenshot because formatting gets messed up in a discuss text post)

• Antoine Dahan

Here is a much simpler solution:

• Chill

The result gives 1->2->2->4->3->5. It is wrong.
It has to be 1->2->2->3->4->5.

Rule: all nodes less than x come before nodes greater than or equal to x.

Please correct me if I am wrong

• Parag Chaudhari

Here is another version,

public void partitionList(Node start, int num){
Node p = start;
Node q = start;
while(q! = null){
while(p != null && p.value num)
q = q.next;
swapValue(p,q);
}
}