Partition List - LeetCode

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; *         next = null; *     } * } */public class Solution {    public ListNode partition(ListNode head, int x) {        if(head == null){        return null;        }        ListNode big = new ListNode(0);        ListNode bigcurrent = new ListNode(0);        big = bigcurrent;        ListNode small = new ListNode(0);        ListNode smallhead = new ListNode(0);        smallhead = small;        ListNode current = new ListNode(0);        current = head;        while(current != null){            if(current.val < x){                small.next = new ListNode(current.val);                small = small.next;            }            else{                bigcurrent.next =  new ListNode(current.val);                bigcurrent = bigcurrent.next;                }            current = current.next;        }        small.next = big.next;        return smallhead.next;    }}