Leetcode刷题笔记1 链表part01

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Leetcode203. 移除链表元素

此题用来熟悉链表结构以及链表常规操作

python中链表定义：

# Definition for singly-linked list.
class ListNode:
    def __init__(self, val=0, next=None):
        self.val = val
        self.next = next

移除链表元素很简单，即将节点跳过所要删除的节点指向下下个节点：

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

Leetcode 707 设计链表

通过此题锻炼对链表的基础操作，但是中间卡了bug耽搁了时间

class ListNode:
    def __init__(self, _val = 0, _next = None):
        self.val = _val
        self.next = _next 

class MyLinkedList:
    def __init__(self):
        self.head = None
        self.size = 0

    def get(self, index: int) -> int:
        if index < 0 or index >= self.size:
            return -1
        tmp = self.head
        for _ in range(index):
            tmp = tmp.next
        return tmp.val

    def addAtHead(self, val: int) -> None:
        self.head = ListNode(val,self.head)
        self.size += 1

    def addAtTail(self, val: int) -> None:
        self.addAtIndex(self.size,val)

    def addAtIndex(self, index: int, val: int) -> None:
        if index > self.size:
            return
        if index <= 0:
            self.addAtHead(val)
            return
        tmp = self.head
        for _ in range(index - 1):
            tmp = tmp.next
        tmp.next = ListNode(val,tmp.next)
        self.size += 1


    def deleteAtIndex(self, index: int) -> None:
        if index < 0 or index >= self.size:
            return
        if index == 0:
            self.head = self.head.next
        else:
            tmp = self.head 
            for _ in range(index - 1):
                tmp = tmp.next
            tmp.next = tmp.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)

Leetcode206 反转链表

这题比较简单，先声明一个空的节点pre，再存一下tmp=cur.next,再将cur.next = pre，pre= cur，cur = tmp

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