当前位置：首页 > news >正文

力扣每日一题2022-09-23中等题：设计链表

news 来源：原创 2024/5/10 7:05:40

设计链表

- 题目描述
- 思路
- - 单链表
  - - Python实现
    - Java实现
    - C++实现
  - 双链表
  - - Python实现

题目描述

设计链表

思路

单链表

根据题意，可以选择单链表和双链表解题，先使用单链表。
因为使用单链表，即每个节点只存储值和直接后继节点，还需要一个哨兵节点（sentinel node）作为头节点，和一个size参数保存有效节点数。
初始化时，只需创建头节点head和size即可。
实现get(index)时，先判断有效性，再通过循环找到对应节点的值。
实现addAtIndex(index, val)时，如果index是有效值，则需要找到原来下标为index的节点的直接前驱结点pred，并创建新节点cur，将cur的直接后继节点设为pred的直接后继节点，将pred的直接后继节点设为cur，更新size，这样的操作对于index=0也成立。
实现addAtHead(val)和addAtTail(val)，可以借助addAtIndex(index, val)实现。
实现deleteAtIndex(index)时，需要先判断index是有效值。然后找到下标为index的节点的直接前驱节点pred，通过将pred的直接后继节点更新为pred的直接后继节点的直接后继节点，来大城删除节点的效果，同时更新size。

Python实现

class ListNode:

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


class MyLinkedList:

    def __init__(self):
        self.size = 0
        self.head = ListNode(0)


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


    def addAtHead(self, val: int) -> None:
        self.addAtIndex(0, val)


    def addAtTail(self, val: int) -> None:
        self.addAtIndex(self.size, val)


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


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

Java实现

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

class MyLinkedList {
    
    int size;
    ListNode head;

    public MyLinkedList() {
        size = 0;
        head = new ListNode(0);
    }
    
    public int get(int index) {
        if (index < 0 || index >= size) {
            return -1;
        }
        ListNode cur = head;
        for (int i = 0; i <= index; i++) {
            cur = cur.next;
        }
        return cur.val;
    }
    
    public void addAtHead(int val) {
        addAtIndex(0, val);
    }
    
    public void addAtTail(int val) {
        addAtIndex(size, val);
    }
    
    public void addAtIndex(int index, int val) {
        if (index > size) {
            return;
        }
        index = Math.max(0, index);
        size++;
        ListNode pred = head;
        for (int i = 0; i < index; i++) {
            pred = pred.next;
        }
        ListNode cur = new ListNode(val);
        cur.next = pred.next;
        pred.next = cur;
    }
    
    public void deleteAtIndex(int index) {
        if (index < 0 || index >= size) {
            return;
        }
        size--;
        ListNode pred = head;
        for (int i = 0; i < index; i++) {
            pred = pred.next;
        }
        pred.next = pred.next.next;
    }
}

/**
 * Your MyLinkedList object will be instantiated and called as such:
 * MyLinkedList obj = new MyLinkedList();
 * int param_1 = obj.get(index);
 * obj.addAtHead(val);
 * obj.addAtTail(val);
 * obj.addAtIndex(index,val);
 * obj.deleteAtIndex(index);
 */

C++实现

class MyLinkedList {
private:
    int size;
    ListNode *head;

public:
    MyLinkedList() {
        this -> size = 0;
        this->head = new ListNode(0);
    }
    
    int get(int index) {
        if (index < 0 || index >= size) {
            return -1;
        }
        ListNode *cur = head;
        for (int i = 0; i <= index; i++) {
            cur = cur->next;
        }
        return cur->val;
    }
    
    void addAtHead(int val) {
        addAtIndex(0, val);
    }
    
    void addAtTail(int val) {
        addAtIndex(size, val);
    }
    
    void addAtIndex(int index, int val) {
        if (index > size) {
            return;
        }
        index = max(0, index);
        size++;
        ListNode *pred = head;
        for (int i = 0; i < index; i++) {
            pred = pred->next;
        }
        ListNode *cur = new ListNode(val);
        cur->next = pred->next;
        pred->next = cur;
    }
    
    void deleteAtIndex(int index) {
        if (index < 0 || index >= size) {
            return;
        }
        size--;
        ListNode *pred = head;
        for (int i = 0; i < index; i++) {
            pred = pred->next;
        }
        ListNode *cur = pred->next;
        pred->next = pred->next->next;
        delete cur;
    }
};

/**
 * Your MyLinkedList object will be instantiated and called as such:
 * MyLinkedList* obj = new MyLinkedList();
 * int param_1 = obj->get(index);
 * obj->addAtHead(val);
 * obj->addAtTail(val);
 * obj->addAtIndex(index,val);
 * obj->deleteAtIndex(index);
 */

双链表

题目给定的另一种方法是双链表，实际上就是每个节点多存储一个直接前驱节点，除此之外，哨兵节点共需要两个，分别为头节点和尾节点，还需要一个size计算有效节点数。
初始化时，创建头节点和size即可。
实现get(index)时，先判断有效性，比较看是从head开始还是从tail开始遍历会更快找到目标，然后开始遍历。
实现addAtIndex(index, val)时，如果index是有效值，则找到原来下标为index的节点next和其直接前驱节点pred，并创建新节点cur，再对pred和next节点变量地更新来插入节点cur，最后更新size。
addAtHead(val)和addAtTail(val)都可以通过addAtIndex(index, val)来实现。
实现deleteAtIndex(index)时，先判断index有效性，然后找到下标为index的节点cur和其直接前驱节点pred和直接后继节点next，再对pred和next的变量进行修改来删除节点，最后更新size。

Python实现

class ListNode:

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


class MyLinkedList:

    def __init__(self):
        self.size = 0
        self.head, self.tail = ListNode(0), ListNode(0) 
        self.head.next = self.tail
        self.tail.prev = self.head


    def get(self, index: int) -> int:
        if index < 0 or index >= self.size:
            return -1
        if index + 1 < self.size - index:
            curr = self.head
            for _ in range(index + 1):
                curr = curr.next
        else:
            curr = self.tail
            for _ in range(self.size - index):
                curr = curr.prev
        return curr.val


    def addAtHead(self, val: int) -> None:
        self.addAtIndex(0, val)


    def addAtTail(self, val: int) -> None:
        self.addAtIndex(self.size, val)


    def addAtIndex(self, index: int, val: int) -> None:
        if index > self.size:
            return
        index = max(0, index)
        if index < self.size - index:
            pred = self.head
            for _ in range(index):
                pred = pred.next
            succ = pred.next
        else:
            succ = self.tail
            for _ in range(self.size - index):
                succ = succ.prev
            pred = succ.prev
        self.size += 1
        to_add = ListNode(val)
        to_add.prev = pred
        to_add.next = succ
        pred.next = to_add
        succ.prev = to_add


    def deleteAtIndex(self, index: int) -> None:
        if index < 0 or index >= self.size:
            return
        if index < self.size - index:
            pred = self.head
            for _ in range(index):
                pred = pred.next
            succ = pred.next.next
        else:
            succ = self.tail
            for _ in range(self.size - index - 1):
                succ = succ.prev
            pred = succ.prev.prev
        self.size -= 1
        pred.next = succ
        succ.prev = pred



# 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)