1.删除特点节点

给你一个链表的头节点 head 和一个整数 val ，请你删除链表中所有满足 Node.val == val 的节点，并返回 新的头节点 。

示例 1：

输入：head = [1,2,6,3,4,5,6], val = 6
输出：[1,2,3,4,5]

示例 2：

输入：head = [], val = 1
输出：[]

示例 3：

输入：head = [7,7,7,7], val = 7
输出：[]

Java实现：

    public static ListNode removeElements(ListNode head, int val) {
        ListNode dummyHead = new ListNode(0);
        dummyHead.next = head;
        ListNode temp = dummyHead;
        while (temp.next != null) {
            if (temp.next.val == val) {
                temp.next = temp.next.next;
            } else {
                temp = temp.next;
            }
        }
        return dummyHead.next;
    }

Python实现

    def removeElements(self, head, val):
        while head and head.val == val:
            head = head.next
        if head is None:
            return head
        node = head
        while node.next:
            if node.next.val == val:
                node.next = node.next.next
            else:
                node = node.next
        return head

2.删除倒数第n个节点

给你一个链表，删除链表的倒数第 n 个结点，并且返回链表的头结点。

示例 1：

输入：head = [1,2,3,4,5], n = 2
输出：[1,2,3,5]

示例 2：

输入：head = [1], n = 1
输出：[]

示例 3：

输入：head = [1,2], n = 1
输出：[1]

Java实现：双指针方法

    public static ListNode removeNthFromEndByTwoPoints(ListNode head, int n) {
        ListNode dummy = new ListNode(0);
        dummy.next = head;
        ListNode first = head;
        ListNode second = dummy;
        for (int i = 0; i < n; ++i) {
            first = first.next;
        }
        while (first != null) {
            first = first.next;
            second = second.next;
        }
        second.next = second.next.next;
        ListNode ans = dummy.next;
        return ans;
    }

python实现

    def removeNthFromEnd2(self, head, n):
        dummy = ListNode(0, head)
        fast = head
        slow = dummy
        for i in range(n):
            fast = fast.next

        while fast:
            fast = fast.next
            slow = slow.next

        slow.next = slow.next.next
        return dummy.next

3.删除重复元素

3.1重复元素保留一个

java实现

    public static ListNode deleteDuplicate(ListNode head) {
        if (head == null) {
            return head;
        }
        ListNode cur = head;
        while (cur.next != null) {
            if (cur.val == cur.next.val) {
                cur.next = cur.next.next;
            } else {
                cur = cur.next;
            }
        }
        return head;
    }

python实现

    def deleteDuplicates(self, head):
        if not head:
            return head
        cur = head
        while cur.next:
            if cur.val == cur.next.val:
                cur.next = cur.next.next
            else:
                cur = cur.next
        return head

3.2重复元素不要

Java实现

    public static ListNode deleteDuplicates(ListNode head) {
        if (head == null) {
            return head;
        }

        ListNode dummy = new ListNode(0);
        dummy.next = head;
        ListNode cur = dummy;
        while (cur.next != null && cur.next.next != null) {
            if (cur.next.val == cur.next.next.val) {
                int x = cur.next.val;
                while (cur.next != null && cur.next.val == x) {
                    //删除重复元素
                    cur.next = cur.next.next;
                }
            } else {
                cur = cur.next;
            }
        }
        return dummy.next;
    }

python实现

    def deleteDuplicates2(self, head):
        if not head:
            return head

        dummy = ListNode(0, head)
        cur = dummy
        while cur.next and cur.next.next:
            if cur.next.val == cur.next.next.val:
                x = cur.next.val
                while cur.next and cur.next.val == x:
                    cur.next = cur.next.next
            else:
                cur = cur.next

        return dummy.next