Partition List

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.

题意：给定一个链表和一个目标值X，使得链表一分为二，即小于X的所有节点都放在大于或等于X的节点之前，并且保留节点原有的相对顺序不变

解题思路：使用两个dummy节点，遍历原链表时，遇到小于X的节点，将其存储在dummy1中，遇到大于等于X的节点，将其存储在dummy2中。最后将dummy2链接到dummy1后。返回dummy1.next

/** * 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) {               //存储小于目标值的节点    ListNode dummy1 = new ListNode(0);    ListNode head1 = dummy1;        ListNode dummy2 = new ListNode(0);//存储大于目标值的节点        ListNode head2 = dummy2;        ListNode curNode = head;        while (curNode != null) {        ListNode nextNode = curNode.next;if (curNode.val < x) {head1.next = curNode;head1 = head1.next;head1.next = null;}else {head2.next =curNode;head2 = head2.next;head2.next = null;}curNode = nextNode;}        head1.next = dummy2.next;        return dummy1.next;    }}