设计链表python(leetcode707)

#707. 设计链表

#设计链表的实现。链表中的节点应该具有两个属性：val 和 next

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

class MyLinkedList:

    def __init__(self):
        self.head = Node(0)
        self.length  = 0
    def get(self, index: int) -> int:
    #获取链表第index个元素的val,index无效则返回-1

        if index < 0 or index >= self.length: #注意最后一个节点没有val
            return -1
        
        curr = self.head
        # index steps needed 
        # to move from sentinel node to wanted index
        for _ in range(index + 1):
            curr = curr.next
        return curr.val

    def addAtHead(self, val: int) -> None:
    #在第一个元素之前添加一个值为val的节点，新节点为head节点
        self.addAtIndex(0, val)

    def addAtTail(self, val: int) -> None:
    #将值为val的节点追加为最后一个元素
        self.addAtIndex(self.length, val)

    def addAtIndex(self, index: int, val: int) -> None:
    #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
        if index < 0:
            index = 0
        elif index > self.length:
            return
    
        add_node = Node(val)
        pre = self.head
        for i in range(index):
            pre = pre.next
        add_node.next = pre.next
        pre.next = add_node

        self.length += 1

    def deleteAtIndex(self, index: int) -> None:
    #如果索引 index 有效，则删除链表中的第 index 个节点
        if index < 0 or index >= self.length: # 最后一个节点是null，没有val
            return
        
        self.length -= 1
        # find predecessor of the node to be deleted
        pred = self.head
        for _ in range(index):
            pred = pred.next
            
        # delete pred.next 
        pred.next = pred.next.next
    
# 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)