代码随想录第三天 链表

第一题 https://leetcode.cn/problems/remove-linked-list-elements/submissions/691848826/

# Definition for singly-linked list.
class ListNode:
    def __init__(self, val=0, next=None):
        self.val = val
        self.next = next


from typing import Optional


class Solution:
    def removeElements(self, head: Optional[ListNode], val: int) -> Optional[ListNode]:
        dummy = ListNode(0, head)
        cur = dummy

        while cur.next:
            if cur.next.val == val:
                cur.next = cur.next.next
            else:
                cur = cur.next

        return dummy.next

注意cur.next为待删除目标时cur不往后移，因为后面还可能是相同元素。虚拟头结点还是挺方便的。

第二题 反转链表https://leetcode.cn/problems/reverse-linked-list/description/

    def reverseList(self, head: Optional[ListNode]) -> Optional[ListNode]:
        cur = head
        pre = None
        while cur:
            next = cur.next
            cur.next = pre
            pre = cur
            cur = next
        return pre

代码不难，也用不着虚拟头结点。

第三题 https://leetcode.cn/problems/design-linked-list/description/

设计类的还是有点难度的，注意索引是否越界要提前判断，首尾插入都应该是按索引插入的特例