1. JZ6 从尾到头打印链表

输入：
{1,2,3}
返回值：
[3,2,1]

public ArrayList<Integer> printListFromTailToHead(ListNode listNode) {
        //用来存储链表中节点的值。
        Stack<Integer> reverse = new Stack<>();
        while(listNode != null){
            reverse.push(listNode.val);
            listNode = listNode.next;
        }
        //创建的题目要求的数据类型来存储反向的节点值。
        ArrayList<Integer> result = new ArrayList<>(); 
        while(!reverse.isEmpty()){
            //将值从栈中弹出，并添加到ArrayList中
            result.add(reverse.pop());
        }
        return result;
    }

2. 反转链表

public ListNode ReverseList(ListNode head) {
        if (head == null || head.next == null)
        return head;
        ListNode node = ReverseList(head.next);
        head.next.next = head;
        head.next = null;
        return node;
    }

3. 合并两个有序链表

public ListNode Merge(ListNode list1,ListNode list2) {
        if(list1==null) return list2;
        if(list2==null) return list1;
        if(list1.val<list2.val){
            list1.next = Merge(list1.next,list2);
            return list1;
        } else {
            list2.next = Merge(list1,list2.next);
            return list2;
        }
    }

4 两个链表第一个公共节点

	public ListNode FindFirstCommonNode(ListNode pHead1, ListNode pHead2) {
       ListNode l1 = pHead1, l2 = pHead2;
        while(l1 != l2){
            l1 = (l1==null)?pHead2:l1.next;
            l2 = (l2==null)?pHead1:l2.next;
        }
        return l1;
    }

5. 删除链表的某个节点

public ListNode deleteNode (ListNode head, int val) {
       while(head != null && head.val == val){
           head = head.next;
       }
        ListNode prev = head;
        if(prev!=null){
            while(prev.next!=null){
                if(prev.next.val == val){
                    prev.next = prev.next.next;
                }else{
                    prev = prev.next;
                }
            }
        }
        return head;
    }

6 删除链表重复节点

public ListNode deleteDuplication(ListNode head) {
        if(head == null || head.next == null){
            return head;
        }

        //两个循环，用来应付“1-1-2-2-3-3-4-5…”格式的连续重复结点
        while(head != null && head.next != null && head.val == head.next.val){
            while(head != null && head.next != null && head.val == head.next.val){
                head = head.next;
            }
            head = head.next;
        }
        if(head!=null ){
            head.next = deleteDuplication(head.next);
        }
        return head;
    }