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

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode partition(ListNode head, int x) {
        ListNode l3=head;
        int n=0;
        while(l3!=null)
        {
            l3=l3.next;
            n++;
        }
        
        l3=head;
        int[] a=new int[n];
        int[] b=new int[n];
        int small=0,big=0;
        while(l3!=null)
        {
            if(l3.val<x)
            {
                a[small]=l3.val;
                small++;
                l3=l3.next;
            }
            else
            {
                b[big]=l3.val;
                big++;
                l3=l3.next;
            }
        }
        
        for(int i=0;i<small;i++)
        {   
            
            if(i==0)
            {
                l3=new ListNode(a[i]);
                head=l3;
            }
            else 
            {
                l3.next=new ListNode(a[i]);
                l3=l3.next;
            }
        }
        for(int i=0;i<big;i++)
        {   
            if(small==0)
            {
                l3=new ListNode(b[i]);
                head=l3;
                small=1;
                continue;
            }
            
            l3.next=new ListNode(b[i]);
            l3=l3.next;
            
            if(i==big-1)
            {
                l3.next=new ListNode(b[i]);
                l3.next=null;
            }
        }
        return head;
    }
}