代码随想录算法训练营第三天| 203.移除链表元素 707.设计链表206.翻转链表

python 链表的定义：

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

203.移除链表元素

讲解链接

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：[]

思路：

本体主要考察链表的定义，和对链表概念的理解，以及如何用代码将链表移除元素给 表达出来的能力。

代码如下：

（版本一）虚拟头节点法
# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def removeElements(self, head: Optional[ListNode], val: int) -> Optional[ListNode]:
        # 创建虚拟头部节点以简化删除过程
        dummy_head = ListNode(next = head)
        
        # 遍历列表并删除值为val的节点
        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

本题主要考察链表的一些基本操作，增、删、查、改等

个人理解本题无特殊技巧，直接理解代码记住用法即可，一般太基础的东西需要牢记。

详细讲解链接

（版本一）单链表法
class ListNode:
    def __init__(self, val=0, next=None):
        self.val = val
        self.next = next
        
class MyLinkedList:
    def __init__(self):
        self.dummy_head = ListNode()
        self.size = 0

    def get(self, index: int) -> 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: int) -> None:
        self.dummy_head.next = ListNode(val, self.dummy_head.next)
        self.size += 1

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

    def addAtIndex(self, index: int, val: int) -> 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: int) -> 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


# Your MyLinkedList object will be instantiated and called as such:
# obj = MyLinkedList()
# param_1 = obj.get(index)
# obj.addAtHead(val)
# obj.addAtTail(val)
# obj.addAtIndex(index,val)
# obj.deleteAtIndex(index)

206.翻转链表 

讲解链接

思路： 用一个 tmp 节点作为过渡，先将当前节点指向的节点存在 tmp，将当前节点指向pre（前节点），再将当前节点的值赋值给 pre（移动 pre），将 tmp 节点的值赋值给 cur （移动 cur），至此完成一个节点翻转

代码如下：

（版本一）双指针法
# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def reverseList(self, head: ListNode) -> ListNode:
        cur = head   
        pre = None
        while cur:
            temp = cur.next # 保存一下 cur的下一个节点，因为接下来要改变cur->next
            cur.next = pre #反转
            #更新pre、cur指针
            pre = cur
            cur = temp
        return pre

（版本二）递归法
# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def reverseList(self, head: ListNode) -> ListNode:
        return self.reverse(head, None)
    def reverse(self, cur: ListNode, pre: ListNode) -> ListNode:
        if cur == None:
            return pre
        temp = cur.next
        cur.next = pre
        return self.reverse(temp, cur)

注明：以上代码均来自代码随想录。