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。
deleteAtIndex(index)：如果索引 index 有效，则删除链表中的第 index 个节点。
来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/design-linked-list
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。
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1.虚拟头节点
思路：
1.通过维护一个虚拟的头节点dummyHead来操作链表，从而实现添加和删除操作。
public class MyLinkedList {
private class Node {
public int val;
public Node next;
public Node(int val, Node next) {
this.val = val;
this.next = next;
}
public Node(int val) {
this(val, null);
}
public Node() {
this(0, null);
}
}
//使用虚拟头节点
private Node dummyHead;
private int size;
/**
* Initialize your data structure here.
*/
public MyLinkedList() {
dummyHead = new Node();
size = 0;
}
/**
* Get the value of the index-th node in the linked list. If the index is invalid, return -1.
*/
public int get(int index) {
if (index < 0 || index >= size) return -1;
Node cur = dummyHead.next; //头节点
for (int i = 0; i < index; i++) {
cur = cur.next;
}
return cur.val;
}
/**
* judge linked list is empty?
*/
public boolean isEmpty() {
return size == 0;
}
/**
* 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.
*/
public void addAtHead(int val) {
addAtIndex(0, val);
}
/**
* Append a node of value val to the last element of the linked list.
*/
public void addAtTail(int val) {
addAtIndex(size, 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.
*/
public void addAtIndex(int index, int val) {
if (index < 0 || index > size) {
return;
}
Node prev = dummyHead;
for (int i = 0; i < index; i++) {
prev = prev.next;
}
prev.next = new Node(val, prev.next);
size++;
}
/**
* Delete the index-th node in the linked list, if the index is valid.
*/
public void deleteAtIndex(int index) {
if (index < 0 || index >= size) throw new IllegalArgumentException("Add filed, index is illegal");
Node prev = dummyHead;
for (int i = 0; i < index; i++) {
prev = prev.next;
}
Node retNode = prev.next;
prev.next = retNode.next;
retNode.next = null; //loitering object != memory leak
size--;
}
}
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2.常规解法
思路：
通过维护当前节点cur完成相关操作
public class MyLinkedList2 {
private class Node {
private int val;
private Node next;
public Node(int val, Node next) {
this.val = val;
this.next = next;
}
public Node(int val) {
this(val, null);
}
public Node() {
this(0, null);
}
}
private Node head;
private int size;
public MyLinkedList2() {
head = null;
size = 0;
}
public void addAtHead(int val) {
// Node node = new Node(val);
// node.next = head;
// head = node;
head = new Node(val, head);
size ++;
}
public void addAtIndex(int index, int val) {
if (index < 0 || index > size) {
return;
}
if (index == 0) {
addAtHead(val);
} else {
Node prev = head;
for (int i = 0; i < index - 1; i++) {
prev = prev.next;
}
prev.next = new Node(val, prev.next);
size ++;
}
}
public void deleteAtHead() {
head = head.next;
size--;
}
public void deleteAtIndex(int index) {
if (index < 0 || index >= size) {
return;
}
if (index == 0) {
deleteAtHead();
} else {
Node prev = head;
for (int i = 0; i < index - 1; i++) {
prev = prev.next;
}
Node cur = prev.next;
prev.next = cur.next;
cur.next = null;
size--;
}
}
public void addAtTail(int val) {
addAtIndex(size, val);
}
public int get(int index) {
if (index < 0 || index >= size) {
return -1;
}
if (index == 0) {
return head.val;
} else {
Node cur = head.next;
for (int i = 0; i < index - 1; i++) {
cur = cur.next;
}
return cur.val;
}
}
@Override
public String toString() {
StringBuilder res = new StringBuilder();
Node cur = head;
while (cur != null) {
res.append(cur.val + "->");
cur = cur.next;
}
res.append("NULL");
return res.toString();
}
}
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测试用例
public static void main(String[] args) {
MyLinkedList2 myLinkedList2 = new MyLinkedList2();
myLinkedList2.addAtHead(1);
System.out.println(myLinkedList2);
myLinkedList2.addAtTail(3);
System.out.println(myLinkedList2);
myLinkedList2.addAtIndex(1, 2);
System.out.println(myLinkedList2);
myLinkedList2.deleteAtIndex(1);
System.out.println(myLinkedList2);
myLinkedList2.get(1);
System.out.println(myLinkedList2);
}
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结果
不知道测试用例是否有问题，一直无法通过leetcode中最后一个用例。
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源码
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我会随时更新新的算法，并尽可能尝试不同解法，如果发现问题请指正
- Github
