Merge pull request #2818 from donghuanjie/master

707代码规范化,添加部分comments; 203添加了java版本的递归
This commit is contained in:
程序员Carl
2024-12-15 09:14:58 +08:00
committed by GitHub
2 changed files with 94 additions and 65 deletions

View File

@ -337,6 +337,37 @@ public ListNode removeElements(ListNode head, int val) {
```
递归
```java
/**
* 时间复杂度 O(n)
* 空间复杂度 O(n)
* @param head
* @param val
* @return
*/
class Solution {
public ListNode removeElements(ListNode head, int val) {
if (head == null) {
return head;
}
// 假设 removeElements() 返回后面完整的已经去掉val节点的子链表
// 在当前递归层用当前节点接住后面的子链表
// 随后判断当前层的node是否需要被删除如果是就返回
// 也可以先判断是否需要删除当前node但是这样条件语句会比较不好想
head.next = removeElements(head.next, val);
if (head.val == val) {
return head.next;
}
return head;
// 实际上就是还原一个从尾部开始重新构建链表的过程
}
}
```
### Python
```python

View File

@ -422,38 +422,38 @@ void myLinkedListFree(MyLinkedList* obj) {
```Java
//单链表
class ListNode {
int val;
ListNode next;
ListNode(){}
ListNode(int val) {
this.val=val;
}
}
class MyLinkedList {
class ListNode {
int val;
ListNode next;
ListNode(int val) {
this.val=val;
}
}
//size存储链表元素的个数
int size;
//虚拟头结点
ListNode head;
private int size;
//注意这里记录的是虚拟头结点
private ListNode head;
//初始化链表
public MyLinkedList() {
size = 0;
head = new ListNode(0);
this.size = 0;
this.head = new ListNode(0);
}
//获取第index个节点的数值注意index是从0开始的第0个节点就是头结点
//获取第index个节点的数值注意index是从0开始的第0个节点就是虚拟头结点
public int get(int index) {
//如果index非法返回-1
if (index < 0 || index >= size) {
return -1;
}
ListNode currentNode = head;
//包含一个虚拟头节点,所以查找第 index+1 个节点
ListNode cur = head;
//第0个节点是虚拟头节点,所以查找第 index+1 个节点
for (int i = 0; i <= index; i++) {
currentNode = currentNode.next;
cur = cur.next;
}
return currentNode.val;
return cur.val;
}
public void addAtHead(int val) {
@ -473,7 +473,6 @@ class MyLinkedList {
while (cur.next != null) {
cur = cur.next;
}
cur.next = newNode;
size++;
@ -485,55 +484,53 @@ class MyLinkedList {
// 如果 index 等于链表的长度,则说明是新插入的节点为链表的尾结点
// 如果 index 大于链表的长度,则返回空
public void addAtIndex(int index, int val) {
if (index > size) {
if (index < 0 || index > size) {
return;
}
if (index < 0) {
index = 0;
}
size++;
//找到要插入节点的前驱
ListNode pred = head;
ListNode pre = head;
for (int i = 0; i < index; i++) {
pred = pred.next;
pre = pre.next;
}
ListNode toAdd = new ListNode(val);
toAdd.next = pred.next;
pred.next = toAdd;
ListNode newNode = new ListNode(val);
newNode.next = pre.next;
pre.next = newNode;
size++;
}
//删除第index个节点
public void deleteAtIndex(int index) {
if (index < 0 || index >= size) {
return;
}
size--;
//因为有虚拟头节点,所以不用对Index=0的情况进行特殊处理
ListNode pred = head;
//因为有虚拟头节点,所以不用对index=0的情况进行特殊处理
ListNode pre = head;
for (int i = 0; i < index ; i++) {
pred = pred.next;
pre = pre.next;
}
pred.next = pred.next.next;
pre.next = pre.next.next;
size--;
}
}
```
```Java
//双链表
class ListNode{
int val;
ListNode next,prev;
ListNode() {};
ListNode(int val){
this.val = val;
}
}
class MyLinkedList {
class ListNode{
int val;
ListNode next, prev;
ListNode(int val){
this.val = val;
}
}
//记录链表中元素的数量
int size;
private int size;
//记录链表的虚拟头结点和尾结点
ListNode head,tail;
private ListNode head, tail;
public MyLinkedList() {
//初始化操作
@ -541,25 +538,25 @@ class MyLinkedList {
this.head = new ListNode(0);
this.tail = new ListNode(0);
//这一步非常关键否则在加入头结点的操作中会出现null.next的错误
head.next=tail;
tail.prev=head;
this.head.next = tail;
this.tail.prev = head;
}
public int get(int index) {
//判断index是否有效
if(index>=size){
if(index < 0 || index >= size){
return -1;
}
ListNode cur = this.head;
ListNode cur = head;
//判断是哪一边遍历时间更短
if(index >= size / 2){
//tail开始
cur = tail;
for(int i=0; i< size-index; i++){
for(int i = 0; i < size - index; i++){
cur = cur.prev;
}
}else{
for(int i=0; i<= index; i++){
for(int i = 0; i <= index; i++){
cur = cur.next;
}
}
@ -568,24 +565,23 @@ class MyLinkedList {
public void addAtHead(int val) {
//等价于在第0个元素前添加
addAtIndex(0,val);
addAtIndex(0, val);
}
public void addAtTail(int val) {
//等价于在最后一个元素(null)前添加
addAtIndex(size,val);
addAtIndex(size, val);
}
public void addAtIndex(int index, int val) {
//index大于链表长度
if(index>size){
//判断index是否有效
if(index < 0 || index > size){
return;
}
size++;
//找到前驱
ListNode pre = this.head;
for(int i=0; i<index; i++){
ListNode pre = head;
for(int i = 0; i < index; i++){
pre = pre.next;
}
//新建结点
@ -594,22 +590,24 @@ class MyLinkedList {
pre.next.prev = newNode;
newNode.prev = pre;
pre.next = newNode;
size++;
}
public void deleteAtIndex(int index) {
//判断索引是否有效
if(index>=size){
//判断index是否有效
if(index < 0 || index >= size){
return;
}
//删除操作
size--;
ListNode pre = this.head;
for(int i=0; i<index; i++){
ListNode pre = head;
for(int i = 0; i < index; i++){
pre = pre.next;
}
pre.next.next.prev = pre;
pre.next = pre.next.next;
size--;
}
}