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Commit 977b537

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feat: update leetcode and lcci solutions
1 parent ffb91b8 commit 977b537

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11 files changed

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11 files changed

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‎README.md‎

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1. [从尾到头打印链表](./lcof/面试题06.%20从尾到头打印链表/README.md)
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1. [删除链表的节点](./lcof/面试题18.%20删除链表的节点/README.md)
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1. [链表中倒数第 k 个节点](./lcof/面试题22.%20链表中倒数第k个节点/README.md)
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1. [链表中倒数第 k 个节点](./lcci/02.02.Kth%20Node%20From%20End%20of%20List/README.md)
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1. [反转链表](./solution/0200-0299/0206.Reverse%20Linked%20List/README.md)
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1. [环形链表](./solution/0100-0199/0141.Linked%20List%20Cycle/README.md)
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1. [环形链表 II](./solution/0100-0199/0142.Linked%20List%20Cycle%20II/README.md)

‎README_EN.md‎

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@@ -39,6 +39,8 @@ Complete solutions to [LeetCode](https://leetcode-cn.com/problemset/all/), [LCOF
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### Linked List
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1. [Delete Node in a Linked List](./solution/0200-0299/0237.Delete%20Node%20in%20a%20Linked%20List/README_EN.md)
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1. [Kth Node From End of List](./lcci/02.02.Kth%20Node%20From%20End%20of%20List/README_EN.md)
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1. [Reverse Linked List](./solution/0200-0299/0206.Reverse%20Linked%20List/README_EN.md)
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1. [Linked List Cycle](./solution/0100-0199/0141.Linked%20List%20Cycle/README_EN.md)
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1. [Linked List Cycle II](./solution/0100-0199/0142.Linked%20List%20Cycle%20II/README_EN.md)

‎lcci/02.02.Kth Node From End of List/README.md‎

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## 解法
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<!-- 这里可写通用的实现逻辑 -->
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定义 `p`、`q` 指针指向 `head`。
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`p` 先向前走 `k` 步,接着 `p`、`q` 同时向前走,当 `p` 指向 `null` 时,`q` 指向的节点即为链表的倒数第 `k` 个节点。
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<!-- tabs:start -->
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for _ in range(k):
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q = q.next
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while q:
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q = q.next
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p = p.next
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p, q = p.next, q.next
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return p.val
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```
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‎lcci/02.02.Kth Node From End of List/README_EN.md‎

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@@ -48,8 +48,7 @@ class Solution:
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for _ in range(k):
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q = q.next
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while q:
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q = q.next
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p = p.next
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p, q = p.next, q.next
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return p.val
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```
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‎lcci/02.02.Kth Node From End of List/Solution.py‎

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@@ -10,6 +10,5 @@ def kthToLast(self, head: ListNode, k: int) -> int:
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for _ in range(k):
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q = q.next
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while q:
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q = q.next
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p = p.next
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p, q = p.next, q.next
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return p.val

‎solution/0200-0299/0237.Delete Node in a Linked List/README.md‎

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<p>现有一个链表 --&nbsp;head =&nbsp;[4,5,1,9],它可以表示为:</p>
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<p><img alt="" src="https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2019/01/19/237_example.png" style="height: 49px; width: 300px;"></p>
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<p>&nbsp;</p>
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![](./images/237_example.png)
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<p><strong>示例 1:</strong></p>
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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# Definition for singly-linked list.
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# class ListNode:
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# def __init__(self, x):
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# self.val = x
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# self.next = None
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class Solution:
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def deleteNode(self, node):
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"""
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:type node: ListNode
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:rtype: void Do not return anything, modify node in-place instead.
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"""
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node.val = node.next.val
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node.next = node.next.next
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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```
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### **...**
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/**
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* Definition for singly-linked list.
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* public class ListNode {
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* int val;
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* ListNode next;
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* ListNode(int x) { val = x; }
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* }
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*/
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class Solution {
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public void deleteNode(ListNode node) {
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node.val = node.next.val;
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node.next = node.next.next;
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}
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}
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```
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### **JavaScript**
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```javascript
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/**
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* Definition for singly-linked list.
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* function ListNode(val) {
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* this.val = val;
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* this.next = null;
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* }
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*/
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/**
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* @param {ListNode} node
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* @return {void} Do not return anything, modify node in-place instead.
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*/
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var deleteNode = function(node) {
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node.val = node.next.val
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node.next = node.next.next
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};
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```
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<!-- tabs:end -->

‎solution/0200-0299/0237.Delete Node in a Linked List/README_EN.md‎

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<ul>
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<li>The linked list will have at least two elements.</li>
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<li>All of the nodes&#39; values will be unique.</li>
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<li>The given node&nbsp;will not be the tail and it will always be a valid node of the linked list.</li>
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<li>Do not return anything from your function.</li>
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</ul>
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### **Python3**
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```python
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# Definition for singly-linked list.
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# class ListNode:
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# def __init__(self, x):
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# self.val = x
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# self.next = None
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class Solution:
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def deleteNode(self, node):
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"""
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:type node: ListNode
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:rtype: void Do not return anything, modify node in-place instead.
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"""
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node.val = node.next.val
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node.next = node.next.next
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```
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### **Java**
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```java
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```
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### **...**
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/**
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* Definition for singly-linked list.
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* public class ListNode {
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* int val;
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* ListNode next;
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* ListNode(int x) { val = x; }
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* }
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*/
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class Solution {
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public void deleteNode(ListNode node) {
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node.val = node.next.val;
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node.next = node.next.next;
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}
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}
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```
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### **JavaScript**
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```javascript
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/**
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* Definition for singly-linked list.
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* function ListNode(val) {
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* this.val = val;
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* this.next = null;
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* }
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*/
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/**
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* @param {ListNode} node
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* @return {void} Do not return anything, modify node in-place instead.
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*/
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var deleteNode = function(node) {
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node.val = node.next.val
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node.next = node.next.next
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};
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```
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<!-- tabs:end -->
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/**
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* Definition for singly-linked list.
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* public class ListNode {
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* int val;
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* ListNode next;
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* ListNode(int x) { val = x; }
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* }
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*/
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class Solution {
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public void deleteNode(ListNode node) {
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// 保存下一个结点
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ListNode tmp = node.next;
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// 将下个结点的值赋给当前要删除的结点
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node.val = node.next.val;
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node.next = node.next.next;
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// tmp 置为空,让jvm进行垃圾回收
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tmp = null;
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}
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}
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const deleteNode = function(node){
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node.val = node.next.val;
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node.next = node.next.next;
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}
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/**
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* Definition for singly-linked list.
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* function ListNode(val) {
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* this.val = val;
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* this.next = null;
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* }
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*/
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/**
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* @param {ListNode} node
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* @return {void} Do not return anything, modify node in-place instead.
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*/
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var deleteNode = function(node) {
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node.val = node.next.val
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node.next = node.next.next
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};

‎solution/0200-0299/0237.Delete Node in a Linked List/Solution.py‎

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@@ -10,7 +10,5 @@ def deleteNode(self, node):
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:type node: ListNode
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:rtype: void Do not return anything, modify node in-place instead.
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"""
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tmp=node.next
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node.val=tmp.val
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node.next=tmp.next
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node.val = node.next.val
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node.next = node.next.next

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