86. Partition List

Given a linked list and a valuex, partition it such that all nodes less thanxcome before nodes greater than or equal tox.

You should preserve the original relative order of the nodes in each of the two partitions.

For example,
Given1->4->3->2->5->2andx= 3,
return1->2->2->4->3->5.

需要check最后尾巴是否已经封上，否则会导致循环往复。比如rightTail.next = null

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode partition(ListNode head, int x) {
        if (head == null) {
            return head;
        }
        ListNode dummyLeft = new ListNode(0);
        ListNode leftTail = dummyLeft;
        ListNode dummyRight = new ListNode(0);
        ListNode rightTail = dummyRight;
        while (head != null) {
            if (head.val < x) {
                leftTail.next = head;
                leftTail = head;
            } else {
                rightTail.next = head;
                rightTail = head;
            }
            head = head.next;
        }
        leftTail.next = dummyRight.next;
        rightTail.next = null;
        return dummyLeft.next;
    }
}