[LeetCode] Partition List

Problem:

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.

Solution:O(n)

public class Solution {
    public ListNode partition(ListNode head, int x) {
        ListNode dummys = new ListNode(0);
        ListNode dummyb = new ListNode(0);
        ListNode curs = dummys;
        ListNode curb = dummyb;
        while(head!=null){
            if(head.val <x){
                curs.next = head;
                curs = curs.next;
            }else{
                curb.next = head;
                curb = curb.next;
            }   
            head = head.next;
        }
        curs.next = dummyb.next;
        curb.next = null; // be careful this
        return dummys.next;
    }
}