链表 - 707. 设计链表 - 《LeetCode刷题》

题目
题解

题目

707. 设计链表
设计链表的实现。您可以选择使用单链表或双链表。单链表中的节点应该具有两个属性：val 和 next。val 是当前节点的值，next 是指向下一个节点的指针/引用。如果要使用双向链表，则还需要一个属性 prev 以指示链表中的上一个节点。假设链表中的所有节点都是 0-index 的。

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。

示例：**

MyLinkedList linkedList = new MyLinkedList();
linkedList.addAtHead(1);
linkedList.addAtTail(3);
linkedList.addAtIndex(1,2);   //链表变为1-> 2-> 3
linkedList.get(1);            //返回2
linkedList.deleteAtIndex(1);  //现在链表是1-> 3
linkedList.get(1);            //返回3

提示：**

所有val值都在 [1, 1000] 之内。
操作次数将在 [1, 1000] 之内。
请不要使用内置的 LinkedList 库。

题解

感觉题目不难，就是写起来比较费功夫，中间还忘记更改 self.len 了，结果死活查不出bug。。。

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


class MyLinkedList:
    def __init__(self):
        """
        Initialize your data structure here.
        """
        self.head = None
        self.len = 0

    def get(self, index: int) -> int:
        """
        Get the value of the index-th node in the linked list. If the index is invalid, return -1.
        """
        if index < 0 or index > self.len-1:
            return -1
        curr = self.head
        for i in range(index):
            curr = curr.next
        return curr.val

    def addAtHead(self, val: int) -> None:
        """
        Add a node of value val before the first element of the linked list. After the insertion, the new node will be the first node of the linked list.
        """
        p = ListNode(val)
        p.next = self.head
        self.head = p
        self.len += 1

    def addAtTail(self, val: int) -> None:
        """
        Append a node of value val to the last element of the linked list.
        """
        p = ListNode(val)
        if self.head:
            curr = self.head
            while curr.next:
                curr = curr.next
            curr.next = p
        else:
            self.head = p
        self.len += 1

    def addAtIndex(self, index: int, val: int) -> None:
        """
        Add a node of value val before the index-th node in the linked list. If index equals to the length of linked list, the node will be appended to the end of linked list. If index is greater than the length, the node will not be inserted.
        """
        if index < 0 or index > self.len:
            return
        self.len += 1
        if index == 0:
            self.addAtHead(val)
        elif index == self.len:
            self.addAtTail(val)
        else:
            pre = self.head
            for i in range(index-1):
                pre = pre.next
            curr = ListNode(val)
            curr.next = pre.next
            pre.next = curr

    def deleteAtIndex(self, index: int) -> None:
        """
        Delete the index-th node in the linked list, if the index is valid.
        """
        if index < 0 or index > self.len-1:
            return
        self.len -= 1
        if index == 0:
            self.head = self.head.next
        else:
            pre = self.head
            for i in range(index-1):
                pre = pre.next
            pre.next = pre.next.next

    def travel(self):
        curr = self.head
        print("#"*10+" Traversal Start "+"#"*10)
        while curr:
            print(curr.val)
            curr = curr.next
        print('')


# Your MyLinkedList object will be instantiated and called as such:
obj = MyLinkedList()
obj.addAtHead(1)
obj.travel()
obj.addAtTail(3)
obj.travel()
obj.addAtIndex(1, 2)
obj.travel()
print(obj.get(1))
obj.deleteAtIndex(1)
print(obj.get(1))

后面看了官方的标准答案，用了哨兵节点作为伪头，省去了很多边界情况的讨论，非常机智。

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

class MyLinkedList:
    def __init__(self):
        self.size = 0
        self.head = ListNode(0)  # sentinel node as pseudo-head


    def get(self, index: int) -> int:
        """
        Get the value of the index-th node in the linked list. If the index is invalid, return -1.
        """
        # if index is invalid
        if index < 0 or index >= self.size:
            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:
        """
        Add a node of value val before the first element of the linked list. After the insertion, the new node will be the first node of the linked list.
        """
        self.addAtIndex(0, val)


    def addAtTail(self, val: int) -> None:
        """
        Append a node of value val to the last element of the linked list.
        """
        self.addAtIndex(self.size, val)


    def addAtIndex(self, index: int, val: int) -> None:
        """
        Add a node of value val before the index-th node in the linked list. If index equals to the length of linked list, the node will be appended to the end of linked list. If index is greater than the length, the node will not be inserted.
        """
        # If index is greater than the length, 
        # the node will not be inserted.
        if index > self.size:
            return

        # [so weird] If index is negative, 
        # the node will be inserted at the head of the list.
        if index < 0:
            index = 0

        self.size += 1
        # find predecessor of the node to be added
        pred = self.head
        for _ in range(index):
            pred = pred.next

        # node to be added
        to_add = ListNode(val)
        # insertion itself
        to_add.next = pred.next
        pred.next = to_add


    def deleteAtIndex(self, index: int) -> None:
        """
        Delete the index-th node in the linked list, if the index is valid.
        """
        # if the index is invalid, do nothing
        if index < 0 or index >= self.size:
            return

        self.size -= 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

作者：LeetCode
链接：https://leetcode-cn.com/problems/design-linked-list/solution/she-ji-lian-biao-by-leetcode/
来源：力扣（LeetCode）
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