代码随想录算法训练营|Day 3

Day 3

今日任务

●  链表理论基础

●  203.移除链表元素

●  707.设计链表

●  206.反转链表

详细布置

链表理论基础

建议：了解一下链表基础，以及链表和数组的区别

文章链接：https://programmercarl.com/链表理论基础.html

203.移除链表元素

建议： 本题最关键是要理解 虚拟头结点的使用技巧，这个对链表题目很重要。

题目链接/文章讲解/视频讲解：：https://programmercarl.com/0203.移除链表元素.html

没有虚拟头节点：

点击查看代码

# 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]:
        while head != None and head.val == val:
            head = head.next
        if head == None or head.next == None:
            return head
        p = head
        while p.next != None:
            value = p.next.val
            if value == val:
                p.next = p.next.next
            else:
                p = p.next
        return head

设置dummy nodes：

点击查看代码

# 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

递归的思路解决本题:
基础情况：对于空链表，不需要移除元素。
递归情况：首先检查头节点的值是否为 val，如果是则移除头节点，答案即为在头节点的后续节点上递归的结果；如果头节点的值不为 val，则答案为头节点与在头节点的后续节点上递归得到的新链表拼接的结果。

点击查看代码

class Solution {
public:
    ListNode* removeElements(ListNode* head, int val) {
        // 基础情况：空链表
        if (head == nullptr) {
            return nullptr;
        }

        // 递归处理
        if (head->val == val) {
            ListNode* newHead = removeElements(head->next, val);
            delete head;
            return newHead;
        } else {
            head->next = removeElements(head->next, val);
            return head;
        }
    }
};

707.设计链表

题目链接/文章讲解/视频讲解：https://programmercarl.com/0707.设计链表.html

单链表法

点击查看代码

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
        curr = self.dummy_head.next
        for _ in range(index):
            curr = curr.next
        return curr.val
    def addAtHead(self, val: int) -> None:
        self.addAtIndex(0,val)
        
    def addAtTail(self, val: int) -> None:
        self.addAtIndex(self.size,val)
        
    def addAtIndex(self, index: int, val: int) -> None:
        if index < 0 or index > self.size:
            return  
        curr = self.dummy_head
        for _ in range(index):
            curr = curr.next
        curr.next = ListNode(val, curr.next)
        self.size += 1

    def deleteAtIndex(self, index: int) -> None:
        if index < 0 or index >= self.size:
            return
        curr = self.dummy_head
        for _ in range(index):
            curr = curr.next
        curr.next = curr.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)

双链表法

点击查看代码

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

class MyLinkedList:
    def __init__(self):
        self.head = None
        self.tail = None
        self.size = 0

    def get(self, index: int) -> int:
        if index < 0 or index >= self.size:
            return -1
        
        if index < self.size // 2:
            current = self.head
            for i in range(index):
                current = current.next
        else:
            current = self.tail
            for i in range(self.size - index - 1):
                current = current.prev
                
        return current.val

    def addAtHead(self, val: int) -> None:
        new_node = ListNode(val, None, self.head)
        if self.head:
            self.head.prev = new_node
        else:
            self.tail = new_node
        self.head = new_node
        self.size += 1

    def addAtTail(self, val: int) -> None:
        new_node = ListNode(val, self.tail, None)
        if self.tail:
            self.tail.next = new_node
        else:
            self.head = new_node
        self.tail = new_node
        self.size += 1

    def addAtIndex(self, index: int, val: int) -> None:
        if index < 0 or index > self.size:
            return
        
        if index == 0:
            self.addAtHead(val)
        elif index == self.size:
            self.addAtTail(val)
        else:
            if index < self.size // 2:
                current = self.head
                for i in range(index - 1):
                    current = current.next
            else:
                current = self.tail
                for i in range(self.size - index):
                    current = current.prev
            new_node = ListNode(val, current, current.next)
            current.next.prev = new_node
            current.next = new_node
            self.size += 1

    def deleteAtIndex(self, index: int) -> None:
        if index < 0 or index >= self.size:
            return
        
        if index == 0:
            self.head = self.head.next
            if self.head:
                self.head.prev = None
            else:
                self.tail = None
        elif index == self.size - 1:
            self.tail = self.tail.prev
            if self.tail:
                self.tail.next = None
            else:
                self.head = None
        else:
            if index < self.size // 2:
                current = self.head
                for i in range(index):
                    current = current.next
            else:
                current = self.tail
                for i in range(self.size - index - 1):
                    current = current.prev
            current.prev.next = current.next
            current.next.prev = current.prev
        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.反转链表

题目链接/文章讲解/视频讲解：https://programmercarl.com/0206.翻转链表.html

双指针法：

点击查看代码

# 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: Optional[ListNode]) -> Optional[ListNode]:
        pre = None
        cur = head
        while cur:
            tmp = cur.next
            cur.next = pre
            pre = cur
            cur = tmp
        return pre

递归法：

理解：每次处理一个节点的反转，链表的剩余部分的反转由递归函数处理

点击查看代码

# 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 #保留下个节点避免失去
        cur.next = pre  #反转
        return self.reverse(temp, cur)