203.移除链表元素

创建虚拟头节点比较好做，不然头节点不好处理。这道题很简单，碰到相同的就跳过去了

# 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

先不明白的画一画就会了

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

206.反转链表

• 时间复杂度: O(n)
• 空间复杂度: O(1)

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

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)