203. Remove Linked List Elements

问题
Given the head of a linked list and an integer val, remove all the nodes of the linked list that has Node.val == val, and return the new head.

设置哨兵节点，将其next指向头部。
设置前节点，将其指向哨兵节点。
设置尾部节点，并指向头部。
移动当前节点尾部，如尾部的val等于需要删去的val，则将前节点的next指向尾部的next。
尾部的next如为null，则前节点的next指向null。

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode removeElements(ListNode head, int val) {
        if (head == null){
            return head;
        }
        ListNode dummyHead = new ListNode();
        dummyHead.next = head;
        
        ListNode preNode = dummyHead;
        ListNode tail = dummyHead.next;
        
        while( tail != null ){
            if ( tail.next == null && tail.val == val ){
                preNode.next = null;
                break;
            }
            else if (tail.val == val){
                preNode.next = tail.next;
                tail = preNode.next;
            }
            else{
                preNode = preNode.next;
                tail = tail.next;
            }
        }
        
        return dummyHead.next;
    }
}