Leetcode刷题第十五天-链表

203：移除链表元素

链接：203. 移除链表元素 - 力扣（LeetCode）

# 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]:
        #虚拟头节点
        dummhead=ListNode(next = head)
        cur=dummhead
        while cur.next:
            if(cur.next.val==val):  cur.next=cur.next.next
            else:   cur=cur.next
        return dummhead.next
        '''
        #原链表删除
        #如果头节点等于val，删头节点，一直到头节点不是val
        while head and head.val==val:  head=head.next
        if not head :   return head
        cur=head
        while cur and cur.next:
            if(cur.next.val==val):  cur.next=cur.next.next
            else:   cur=cur.next
        return head
        '''
        

707：设计链表

链接：707. 设计链表 - 力扣（LeetCode）

class ListNode:
    def __init__(self,val=0,next=None):
        self.val=val
        self.next=next
class MyLinkedList:
    
    def __init__(self):
        #虚拟头节点，链表长度为0
        self.dummyhead=ListNode()
        self.size=0

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

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

    def addAtTail(self, val: int) -> None:
        cur=self.dummyhead
        while cur.next: cur=cur.next
        cur.next=ListNode(val)
        self.size+=1

    def addAtIndex(self, index: int, val: int) -> None:
        if index<0 or index > self.size:  return 
        cur=self.dummyhead
        for i in range(index):
            cur=cur.next
        cur.next=ListNode(val,cur.next)
        self.size+=1

    def deleteAtIndex(self, index: int) -> None:
        if index<0 or index>=self.size :    return
        cur=self.dummyhead
        for i in range(index):  cur=cur.next
        cur.next=cur.next.next
        self.size-=1

    def display(self):
        #返回一个list,便于debug
        cur=self.dummyhead
        re=[]
        while cur:
            re.append(cur.val)
            cur=cur.next
        return re


# 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

链接：206. 反转链表 - 力扣（LeetCode）

# 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]:
        if not head:    return head
        pre,cur=None,head
        while cur:
            tmp=cur.next
            cur.next=pre
            pre,cur=cur,tmp
        return pre