203.移除链表元素

题目

题意：删除链表中等于给定值 val 的所有节点。

示例 1：  
输入：head = [1,2,6,3,4,5,6], val = 6  
输出：[1,2,3,4,5]

思路

删除链表元素的步骤

• 被删除节点的前一个节点指向被删除节点的下一个节点

代码

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]:
        dummy_head = ListNode(next=head)
        curr =dummy_head

        while curr.next != None:
            
            if curr.next.val == val:
                curr.next = curr.next.next
            else:
                curr = curr.next
        
        return dummy_head.next

707.设计链表

题目

在链表类中实现这些功能：

• get(index)：获取链表中第 index 个节点的值。如果索引无效，则返回-1。
• addAtHead(val)：在链表的第一个元素之前添加一个值为 val 的节点。插入后，新节点将成为链表的第一个节点。
• addAtTail(val)：将值为 val 的节点追加到链表的最后一个元素。
• addAtIndex(index,val)：在链表中的第 index 个节点之前添加值为 val  的节点。如果 index 等于链表的长度，则该节点将附加到链表的末尾。如果 index 大于链表长度，则不会插入节点。如果index小于0，则在头部插入节点。
• deleteAtIndex(index)：如果索引 index 有效，则删除链表中的第 index 个节点。

思路

注意

• 除了get函数，剩余的方法里cur永远不会指向链表的最后一个节点；
• 添加节点时，需要先操作new node 再操作pre node；
• 删除节点时，注意边界条件，index ！= self.size，否则在删除最后一个节点时会出现空指针异常的情况；

代码

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


class MyLinkedList:
    def __init__(self):
        self.dummy_head = Node()
        self.size = 0

    def get(self, index: int) -> int:
        """
        index starts with zero,
        return the value of the node at index
        """
        #  *if not sure about it, assuming index = 0 to see if it returns correct value
        cur = self.dummy_head.next

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

        for _ in range(index):
            cur = cur.next

        return cur.val

    def addAtHead(self, val: int) -> None:
        self.addAtIndex(0, val)

    def addAtTail(self, val: int) -> None:
        self.addAtIndex(self.size, val)
        # or if cur.next is None, then adding a new node at the end

    def addAtIndex(self, index: int, val: int) -> None:
        """
        the reson why the cur = self.dummy_head is that when we want to add/del a node at index,
        we must operate the previous node of the node at index
        """
        cur = self.dummy_head
        if index > self.size:
            return

        for _ in range(index):
            cur = cur.next

        new_node = Node(val, next=cur.next)
        cur.next = new_node
        self.size += 1

    def deleteAtIndex(self, index: int) -> None:
        # ! we need to ensure that cur.next is the node we want to delete
        cur = self.dummy_head

        # the reson why index >= self.size is that when we delete the node, we not only need to operate the previous node,
        # but we also need to operate the node itself at index, such as node.next.next
        if index < 0 or index >= self.size:
            return

        for _ in range(index):
            cur = cur.next

        cur.next = cur.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)

206.反转链表

题目

思路

代码

#
# @lc app=leetcode.cn id=206 lang=python3
#
# [206] 反转链表
#

# @lc code=start
# 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
# @lc code=end