707. Design Linked List

题目

Design your implementation of the linked list. You can choose to use a singly or doubly linked list.
A node in a singly linked list should have two attributes: val and next. val is the value of the current node, and next is a pointer/reference to the next node.
If you want to use the doubly linked list, you will need one more attribute prev to indicate the previous node in the linked list. Assume all nodes in the linked list are 0-indexed.

Implement the MyLinkedList class:

MyLinkedList() Initializes the MyLinkedList object.
int get(int index) Get the value of the index^th node in the linked list. If the index is invalid, return -1.
void addAtHead(int val) 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.
void addAtTail(int val) Append a node of value val as the last element of the linked list.
void addAtIndex(int index, int val) Add a node of value val before the index^th node in the linked list. If index equals the length of the linked list, the node will be appended to the end of the linked list. If index is greater than the length, the node will not be inserted.
void deleteAtIndex(int index) Delete the index^th node in the linked list, if the index is valid.

Example 1:

Input
["MyLinkedList", "addAtHead", "addAtTail", "addAtIndex", "get", "deleteAtIndex", "get"]
[[], [1], [3], [1, 2], [1], [1], [1]]
Output
[null, null, null, null, 2, null, 3]

Explanation
MyLinkedList myLinkedList = new MyLinkedList();
myLinkedList.addAtHead(1);
myLinkedList.addAtTail(3);
myLinkedList.addAtIndex(1, 2);    // linked list becomes 1->2->3
myLinkedList.get(1);              // return 2
myLinkedList.deleteAtIndex(1);    // now the linked list is 1->3
myLinkedList.get(1);              // return 3

Constraints:

0 <= index, val <= 1000
Please do not use the built-in LinkedList library.
At most 2000 calls will be made to get, addAtHead, addAtTail, addAtIndex and deleteAtIndex.

思路

熟悉链表的遍历和插入的操作

使用 dummy node 可以简化操作 head 的插入和删除
使用 size 可以处理 index 越界的情况

解法

# leetcode submit region begin(Prohibit modification and deletion)
class ListNode:
    def __init__(self, val=0, next=None):
        self.val = val
        self.next = next

class MyLinkedList:

    def __init__(self):
        self.dummy = ListNode(0, None)
        self.size = 0
        

    def get(self, index: int) -> int:
        if index < 0 or index >= self.size:
            return -1
        node = self.dummy

        for i in range(index + 1):
            node = node.next

        return node.val
        

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

    def addAtTail(self, val: int) -> None:
        cur = self.dummy

        while cur.next:
            cur = cur.next

        cur.next = ListNode(val, None)
        self.size += 1


        

    def addAtIndex(self, index: int, val: int) -> None:
        if index < 0 or index > self.size:
            return

        cur = self.dummy

        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 >= self.size:
            return

        cur = self.dummy

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

        cur.next = cur.next.next if cur.next else None

        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)
# leetcode submit region end(Prohibit modification and deletion)

复杂度分析

时间复杂度
- addHead、addTail o(1)
- insert、delete、get o(N)
空间复杂度 O(N)

707. Design Linked List ​

题目 ​

思路 ​

解法 ​

复杂度分析 ​

707. Design Linked List

题目

思路

解法

复杂度分析