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.
[code]
public class Solution {
public ListNode partition(ListNode head, int x) {
if(head ==null || head.next==null)return head;
ListNode d1=new ListNode(0), d2=new ListNode(0),p1=d1, p2=d2;
while(head!=null)
{
if(head.val<x)
{
p1.next=head;
p1=head;
}
else
{
p2.next=head;
p2=head;
}
head=head.next;
}
p1.next=d2.next;
p2.next=null;
return d1.next;
}
}
[Thoughts]
忘记p2.next=null