203. 移除链表元素 - 力扣（LeetCode）

之前没接触过链表，直接看讲解

解题思路

有两种，单独处理头节点，或者设置虚拟头节点，让头节点作为普通节点

方法一 使用原链表

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

class Solution(object):
    def removeElements(self, head, val):
        """
        :type head: ListNode
        :type val: int
        :rtype: ListNode
        """
        while head :
            if head.val == val:
                head = head.next
            else:
                break

        current = head

        while current and current.next is not None:
            if current.next.val == val:
                current.next = current.next.next
            else:
                current = current.next

        return head

方法二 虚拟头节点

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

class Solution(object):
    def removeElements(self, head, val):
        """
        :type head: ListNode
        :type val: int
        :rtype: ListNode
        """
        dummy_head = ListNode()
        dummy_head.next = head

        current = dummy_head

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

        return dummy_head.next

707. 设计链表 - 力扣（LeetCode）

单链表

class ListNode():      
    def __init__(self, val=0, next=None):
        self.val = val
        self.next = next

class MyLinkedList(object):

    def __init__(self):
        self.dummy_head = ListNode()
        self.size = 0

    def get(self, index):
        """
        :type index: int
        :rtype: int
        """
        if index < 0 or index >= self.size:
            return -1
        
        current = self.dummy_head.next
        for i in range(index):
            current = current.next
            
        return current.val

    def addAtHead(self, val):
        """
        :type val: int
        :rtype: None
        """
        self.dummy_head.next = ListNode(val, self.dummy_head.next)
        self.size += 1


    def addAtTail(self, val):
        """
        :type val: int
        :rtype: None
        """
        current = self.dummy_head
        while current.next:
            current = current.next
        current.next = ListNode(val)
        self.size += 1

    def addAtIndex(self, index, val):
        """
        :type index: int
        :type val: int
        :rtype: None
        """
        if index < 0 or index > self.size:
            return
        
        current = self.dummy_head
        for i in range(index):
            current = current.next
        current.next = ListNode(val, current.next)
        self.size += 1

    def deleteAtIndex(self, index):
        """
        :type index: int
        :rtype: None
        """
        if index < 0 or index >= self.size:
            return

        current = self.dummy_head
        for i in range(index):
            current = current.next
        current.next = current.next.next
        self.size -= 1

• 下标是从dummy_head.next开始，从0开始
• 参考代码随想录 (programmercarl.com)

206. 反转链表 - 力扣（LeetCode）

双指针法

    def reverseList(self, head):
        """
        :type head: ListNode
        :rtype: ListNode
        """

        cur = head
        pre = None

        while cur:
            temp = cur.next #保存cur下一个节点
            cur.next = pre # 反转

            pre = cur  #面向对象
            cur = temp           

        return pre

递归

class Solution(object):
    def reverseList(self, head):
        """
        :type head: ListNode
        :rtype: ListNode
        """
        return self.reverseNode(head, None)
    
    def reverseNode(self, cur, pre):
        if cur is None:
            return pre

        temp = cur.next
        cur.next = pre
        return self.reverseNode(temp, cur)
               

• 注意类的写法
• 递归还需练习