链表 - Day1

val 为节点值，next 为节点 listnode

创建虚拟节点 会方便删除等操作 dummy = listnode[next = head]

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

## self调用实例，例如data.min()前面的即为实例

创建装饰器（本质上是一个高阶函数，它接受一个函数作为参数，并返回一个新的函数，可以使函数每次都连带执行）

@staticmethod：将方法定义为静态方法，不需要传递 self 参数。

@classmethod：将方法定义为类方法，第一个参数是类对象 cls。

@property：将方法转换为属性，可以通过访问属性的方式调用方法。

from typing import Optional, List
class ListNode:
    def __init__(self, val, next = None):
        self.val = val
        self.next = next

# 辅助函数: 将列表转换成链表
def list_to_linkedlist(elements: List[int]) -> Optional[ListNode]:
    dummy_head = ListNode()
    current = dummy_head
    for element in elements:
        current.next = ListNode(val=element)
        current = current.next
    return dummy_head.next

# 辅助函数: 将链表转换成列表
def linkedlist_to_list(head: Optional[ListNode]) -> List[int]:
    elements = []
    current = head
    while current:
        elements.append(current.val)
        current = current.next
    return elements

# 装饰器: 转换输入和输出
def convert_input_output(func):
    def wrapper(self, head_list: List[int], val: int) -> List[int]:
        head = list_to_linkedlist(head_list)        # 将列表转换成链表
        new_head = func(self, head, val)            # 调用原函数
        result_list = linkedlist_to_list(new_head)  # 将结果链表转换回列表
        return result_list
    return wrapper

203.移除链表元素

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 。

主要在于加虚拟节点，然后依次遍历

class Solution:  
    @convert_input_output
    def removeElements(self, head: Optional[ListNode], val: int) -> Optional[ListNode]:

        dummy_head = ListNode(next = head)              # 创建虚拟头部节点以简化删除过程
        
        current = dummy_head                            # 遍历列表并删除值为val的节点
        while current.next:
            if current.next.val == val:
                current.next = current.next.next
            else:
                current = current.next
        
        return dummy_head.next

707.设计链表的五个接口

获取链表第index个节点的数值

在链表的最前面插入一个节点

在链表的最后面插入一个节点

在链表第index个节点前面插入一个节点

删除链表的第index个节点

class MyLinkedList:

    def __init__(self):
        self.dummy_head = ListNode()
        self.size = 0
        
    def get(self, index: int) -> int:
        current = self.dummy_head
        if index >= 0 and index <= self.size:
            for i in range(index):
                current = current.next
            return current.val
        else:
            return -1

    def addAtHead(self, val: int) -> None:
        new_head = 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:
        current = self.dummy_head

        if index >= 0 and index <= self.size:
            for i in range(index-1):
                current = current.next
            new_node = ListNode(val, current.next)
            current.next = new_node
        else:
            return
        self.size += 1


    def deleteAtIndex(self, index: int) -> None:
        current = self.dummy_head

        if index >= 0 and index <= self.size:
            for i in range(index - 1):
                current = current.next
            current.next = current.next.next
        else:
            return
        self.size -= 1

链表 - Day2

206.反转链表

双指针，类似数组，但要注意的是需要增加临时节点存储原链表下一节点，否则链表转向后会断开无法继续循环

class Solution:
    @convert_input_output
    def reverseList(self, head: Optional[ListNode]) -> Optional[ListNode]:

        cur = head
        pre = None              # 不可以用listnode()，会使得结果增加一个节点

        while cur:
            tmp = cur.next      # 存储下一节点，为了遍历节点
            cur.next = pre      # 反转方向
            pre = cur           # 更新 新链表
            cur = tmp           # 遍历原方向下一个节点
        return pre

24.两两交换链表中的节点

需要递归，循环内交换前后两个节点，前节点向后连接时递归后面的节点

class Solution:
    def swapPairs(self, head: Optional[ListNode]) -> Optional[ListNode]:
        if head is None or head.next is None:
            return head
            
        pre = head
        cur = head.next
        next = head.next.next

        cur.next = pre                      # 后节点连接前节点
        pre.next = self.swapPairs(next)     # 前节点连接后后节点，该节点需更新，递归
        return cur

19.删除链表的倒数第 N 个结点

好牛逼的思路，快指针先走n+1步，剩下走的路就是慢指针需要走的路，就可以找到倒数n的节点位置

class Solution:
    @convert_input_output
    def removeNthFromEnd(self, head: Optional[ListNode], n: int) -> Optional[ListNode]:
  
        dummy = ListNode(next = head)
        slow = fast = dummy

        for i in range(n+1):
            fast = fast.next

        while fast:
            slow = slow.next
            fast = fast.next
        
        slow.next = slow.next.next
        return dummy.next

面试题 02.07. 链表相交

重点在于对齐，因为如果相交，后续节点长度相等，只要从短的一条链同步进行即可，从后往前判断不现实

class Solution:
    def getIntersectionNode(self, headA: ListNode, headB: ListNode) -> ListNode:

        lenA = 0
        lenB = 0
        curA = headA
        curB = headB

        while curA:                    # 计算长度
            lenA += 1
            curA = curA.next
        while curB:
            lenB += 1
            curB = curB.next

        curA = headA                    # 重置，并排序
        curB = headB
        if lenA > lenB:                
            curA, curB = curB, curA
            lenA, lenB = lenB, lenA
        
        for i in range(lenB - lenA):    # 长度对齐
            curB = curB.next

        while curA:
            if curA == curB:
                return curA
            curA = curA.next
            curB = curB.next
        return None      
                    

142.环形链表 II

快的走两步，慢的走一步，环形的话一定会相遇，相遇时重置慢指针

2(x+y) = x+n(y+z)+y

x+y = n(y+z) 原地走n圈

x = n(y+z)-y 即回到相遇点

class Solution:
    def detectCycle(self, head: Optional[ListNode]) -> Optional[ListNode]:
        
        slow = head
        fast = head

        while fast and fast.next:
            slow = slow.next
            fast = fast.next.next
            if slow == fast:
                slow = head
                while slow != fast:
                    slow = slow.next
                    fast = fast.next
                return slow
        
        return None