1. 删除链表节点

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode deleteNode(ListNode head, int val) {
        if (head.val == val) {
            if (head.next == null) return null;
            head = head.next;
            return head;
        }
        ListNode tmp = head;
        while(tmp.next != null) {
            if (tmp.next.val == val) {
                tmp.next = tmp.next.next;
                return head;
            } else {
                tmp = tmp.next;
            }
        }
        return head;
    }
}

2. 删除链表中的重复节点

给定一个排序链表，删除所有含有重复数字的节点，只保留原始链表中 没有重复出现 的数字。

示例 1:
输入: 1->2->3->3->4->4->5
输出: 1->2->5

示例 2:
输入: 1->1->1->2->3
输出: 2->3

解法 1. 递归：相当于从后往前删，记录一个repeatNum ，只要当前的值等于repeatNum ，返回next

class Solution {
    private int repeatNum = Integer.MAX_VALUE;

    public ListNode deleteDuplicates(ListNode head) {
        if(head == null || head.next == null)  return head;
        // 1. 递归中获取正确的 next
        head.next = deleteDuplicates(head.next);
        // 2. head 值等于 next ， 标记 repeatNum
        if(head != null && head.next != null && head.val == head.next.val){
            repeatNum = head.val;
        }
        // 3. head 值为 repeatNum 一律跳过
        while(head != null && head.val == repeatNum){
            head = head.next;
        }

        return head;
    }
}

解法 2. 非递归：使用快慢节点，慢节点作为哨兵节点处理头节点，快节点处理重复节点

class Solution {
    public ListNode deleteDuplicates(ListNode head) {
        if (head == null) return head;
        ListNode dummy = new ListNode(Integer.MIN_VALUE);
        dummy.next = head;
        ListNode slow = dummy;
        ListNode fast = dummy.next;
        while (fast != null) {
            while (fast.next != null && fast.val == fast.next.val) fast = fast.next;
            if (slow.next == fast) {
              // 无重复
              slow = slow.next;
            } else {
              // 有重复，将最后一个重复节点也去除
              slow.next = fast.next;
            }
            fast = fast.next;
        }
        return dummy.next; 
    }
}