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.

一刷
题解：
创建两个新链表，每次当前节点值比x小的放入headA，其余放入headB，最后把两个链表接起来。 要注意把右侧链表的下一节点设置为null。

Time Complexity - O(n)， Space Complexity - O(1)

/**
 * 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) {
        if(head == null || head.next == null)
            return head;
        ListNode headLeft = new ListNode(-1);
        ListNode headRight = new ListNode(-1);
        ListNode nodeLeft = headLeft;
        ListNode nodeRight = headRight;
        
        while(head != null){
            if(head.val < x){
                nodeLeft.next = head;
                nodeLeft = nodeLeft.next;
            } else {
                nodeRight.next = head;
                nodeRight = nodeRight.next;
            }
                
            head = head.next;
        }
        
        nodeRight.next = null;
        nodeLeft.next = headRight.next;
        return headLeft.next;
    }
}