C#LeetCode刷题之#707-设计链表(Design Linked List)

问题

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。
deleteAtIndex(index):如果索引 index 有效,则删除链表中的第 index 个节点。

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。

Design your implementation of the linked list. You can choose to use the singly linked list or the 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 these functions in your linked list class:

get(index) : Get the value of the index-th node in the linked list. If the index is invalid, return -1.
addAtHead(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.
addAtTail(val) : Append a node of value val to the last element of the linked list.
addAtIndex(index, val) : 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.
deleteAtIndex(index) : Delete the index-th node in the linked list, if the index is valid.

MyLinkedList linkedList = new MyLinkedList();

linkedList.addAtHead(1);

linkedList.addAtTail(3);

linkedList.addAtIndex(1, 2);  // linked list becomes 1->2->3

linkedList.get(1);            // returns 2

linkedList.deleteAtIndex(1);  // now the linked list is 1->3

linkedList.get(1);            // returns 3

Note:

All values will be in the range of [1, 1000].
The number of operations will be in the range of [1, 1000].
Please do not use the built-in LinkedList library.

示例

public class Program {

    public static void Main(string[] args) {
        var linkedList = new MyLinkedList();

        linkedList.AddAtHead(1);
        linkedList.AddAtTail(3);
        linkedList.AddAtIndex(1, 2);
        Console.WriteLine(linkedList.Get(1));
        linkedList.DeleteAtIndex(1);
        Console.WriteLine(linkedList.Get(1));

        Console.ReadKey();
    }

    public class ListNode {
        public int val;
        public ListNode next;
        public ListNode(int x) { val = x; }
    }

    public class MyLinkedList {

        private ListNode _listNode = null;
        private int _count = 0;

        public MyLinkedList() {
            _listNode = new ListNode(0);
        }

        public int Get(int index) {
            if(index < 0 || index >= _count) return -1;
            var idx = 0;
            var node = _listNode;
            while(idx++ != index) {
                node = node.next;
            }
            return node.val;
        }

        public void AddAtHead(int val) {
            if(_count == 0) {
                _listNode.val = val;
            } else {
                var newHead = new ListNode(val);
                newHead.next = _listNode;
                _listNode = newHead;
            }
            _count++;
        }

        public void AddAtTail(int val) {
            AddAtIndex(_count, val);
        }

        public void AddAtIndex(int index, int val) {
            if(index < 0 || index > _count) return;
            if(index == 0) {
                var newHead = new ListNode(val);
                newHead.next = _listNode;
                _listNode = newHead;
                _count++;
                return;
            }
            var pre = _listNode;
            var idx = 0;
            while(idx++ != index - 1) {
                pre = pre?.next;
            }
            if(pre == null) return;
            var temp = pre.next;
            pre.next = new ListNode(val);
            pre.next.next = temp;
            _count++;
        }

        public void DeleteAtIndex(int index) {
            if(index < 0 || index >= _count) return;
            var pre = _listNode;
            var idx = 0;
            while(idx++ != index - 1) {
                pre = pre.next;
            }
            pre.next = pre.next.next;
            _count--;
        }
    }

}

以上给出1种算法实现,以下是这个案例的输出结果:

2
3

分析:

这道题在LeetCode为简单,基础知识也很简单,但想要AC真的要花不少功夫,各种边界的处理,另外可以抽出一个独立的Find方法用于查找某索引处的ListNode,各种看官有兴趣的自己尝试一下。

显而易见,Get、AddAtTailAdd、AtIndex、DeleteAtIndex 的时间复杂度为: O(n)​ ,AddAtHead 的时间复杂度为: O(1) 。

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