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/**
 * 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) {
        ListNode root = new ListNode(0);
        ListNode pivot = new ListNode(x);
        ListNode root_next = root, pivot_next = pivot;
        ListNode cur_node = head;
        while(cur_node != null){
            ListNode next = cur_node.next;
            if(cur_node.val >= x){
                pivot_next.next = cur_node;
                pivot_next = cur_node;
                pivot_next.next = null;
            }else{
                root_next.next = cur_node;
                root_next = cur_node;
            }
            cur_node = next;
        }
        root_next.next = pivot.next;
        return root.next;
    }
    public ListNode partitionJiuZhang(ListNode head, int x) {
        if (head == null) {
             return null;
        }
        ListNode leftDummy = new ListNode(0);
        ListNode rightDummy = new ListNode(0);
        ListNode left = leftDummy, right = rightDummy;
        while(head != null) {
            if(head.val < x) {
                left.next = head;
                left = head;
            } else {
                right.next = head;
                right = head;
            }
            head = head.next;
        }
        right.next = null;
        left.next = rightDummy.next;
        return leftDummy.next;
    }
}