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.

思路一：

遍历原链表，将比x小的元素拿出来依次插入新的链表，最后将原链表剩下的节点插在新链表后面即得到结果。

public ListNode partition(ListNode head, int x) {
        //dummy节点用来记录原列表的头
        ListNode dummy = new ListNode(0);
        dummy.next = head;
        ListNode pre = dummy;
        //ret依次存储比x小的节点
        ListNode ret = new ListNode(0);
        //tail节点用来记录ret链表的尾部
        ListNode tail = ret;
        //遍历原链表
        while (head != null) {
            //当前节点值比x小时，插入在tail节点后
            if (head.val < x) {
                tail.next = head;
                tail = tail.next;
                pre.next = head.next;
            } else {
                pre = pre.next;
            }
            head = head.next;
        }
        //将原链表剩下的节点直接加在tail节点后面
        tail.next = dummy.next;
        return ret.next;
    }