代码随想录day3| 203.移除链表元素 707.设计链表 206.反转链表

移除链表元素

（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

设计链表

（版本一）单链表法
class ListNode:
    # 定义链表节点结构体
    def __init__(self, val=0, next=None):
        self.val = val
        self.next = next
        
class MyLinkedList:
    # 初始化链表，
    def __init__(self):
        # 这里定义的头结点 是一个虚拟头结点，而不是真正的链表头结点
        self.dummy_head = ListNode()
        self.size = 0
	
	# 获取到第index个节点数值，如果index是非法数值直接返回-1， 注意index是从0开始的，第0个节点就是头结点
    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
	
	# 在链表最前面插入一个节点，插入完成后，新插入的节点为链表的新的头结点
    def addAtHead(self, val: int) -> None:
        self.dummy_head.next = ListNode(val, self.dummy_head.next)
        self.size += 1
	
	#  在链表最后面添加一个节点
    def addAtTail(self, val: int) -> None:
        current = self.dummy_head
        while current.next:
            current = current.next
        current.next = ListNode(val)
        self.size += 1

    # 在第index个节点之前插入一个新节点，例如index为0，那么新插入的节点为链表的新头节点。
    # 如果index 等于链表的长度，则说明是新插入的节点为链表的尾结点
    # 如果index大于链表的长度，则返回空
    # 如果index小于0，则在头部插入节点
    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
	
	#  删除第index个节点，如果index 大于等于链表的长度，直接return，注意index是从0开始的
    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)

时间复杂度: 涉及 index 的相关操作为 O(index), 其余为 O(1)
空间复杂度: O(n)

反转链表

（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)

思考

需要学会建立虚拟节点的思想