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@TOC

第一题

707. 设计链表 - 力扣（LeetCode）

代码

python版本

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class MyLinkedList:

    def __init__(self):
        self.dummy = ListNode(-1)
        self.size = 0

    def get(self, index: int) -> int:
        if index < 0 or index >= self.size:
            return -1

        cur = self.dummy.next
        i = 0
        while i < index:
            cur = cur.next
            i += 1

        return cur.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:
            index = 0
        elif index > self.size:
            return

        pre = self.dummy
        i = 0
        while i < index:
            pre = pre.next
            i += 1

        pre.next = ListNode(val, pre.next)
        self.size += 1
        return

    def deleteAtIndex(self, index: int) -> None:
        if index < 0 or index >= self.size:
            return

        pre = self.dummy
        i = 0
        while i < index:
            pre = pre.next
            i += 1

        pre.next = pre.next.next
        self.size -= 1
        return

第二题

876. 链表的中间结点 - 力扣（LeetCode）

代码

python版本

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class Solution:
    def middleNode(self, head: Optional[ListNode]) -> Optional[ListNode]:
        slow = fast = head
        while fast and fast.next:
            fast = fast.next.next
            slow = slow.next

        return slow

第三题

203. 移除链表元素 - 力扣（LeetCode）

代码

python版本

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class Solution:
    def removeElements(self, head: Optional[ListNode], val: int) -> Optional[ListNode]:
        dummy = ListNode(-1, head)
        cur = dummy
        while cur.next:
            if cur.next.val == val:
                cur.next = cur.next.next
            else:
                cur = cur.next
        return dummy.next

第四题

83. 删除排序链表中的重复元素 - 力扣（LeetCode）

代码

python版本

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class Solution:
    def deleteDuplicates(self, head: Optional[ListNode]) -> Optional[ListNode]:
        dummy = ListNode(-101, head)
        cur = dummy
        while cur.next:
            if cur.val == cur.next.val:
                cur.next = cur.next.next
            else:
                cur = cur.next
        
        return dummy.next

第五题

19. 删除链表的倒数第 N 个结点 - 力扣（LeetCode）

代码

python版本

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class Solution:
    def removeNthFromEnd(self, head: Optional[ListNode], n: int) -> Optional[ListNode]:
        dummy = ListNode(-1, head)
        slow = fast = dummy

        i = 0
        while i < n:
            fast = fast.next
            i += 1

        while fast.next:
            fast = fast.next
            slow = slow.next

        slow.next = slow.next.next
        return dummy.next