feat: update leetcode and lcci solutions

Author: yanglbme

README.md

@@ -42,7 +42,7 @@
 
 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)
 1. [删除链表的节点](./lcof/面试题18.%20删除链表的节点/README.md)
-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)
+1. [链表中倒数第 k 个节点](./lcci/02.02.Kth%20Node%20From%20End%20of%20List/README.md)
 1. [反转链表](./solution/0200-0299/0206.Reverse%20Linked%20List/README.md)
 1. [环形链表](./solution/0100-0199/0141.Linked%20List%20Cycle/README.md)
 1. [环形链表 II](./solution/0100-0199/0142.Linked%20List%20Cycle%20II/README.md)

README_EN.md

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

@@ -21,6 +21,10 @@
 ## 解法
 <!-- 这里可写通用的实现逻辑 -->
 
+定义 `p`、`q` 指针指向 `head`。
+
+`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。
+
 
 <!-- tabs:start -->
 
@@ -40,8 +44,7 @@ class Solution:
         for _ in range(k):
             q = q.next
         while q:
-            q = q.next
-            p = p.next
+            p, q = p.next, q.next
         return p.val
 ```
 

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

@@ -48,8 +48,7 @@ class Solution:
         for _ in range(k):
             q = q.next
         while q:
-            q = q.next
-            p = p.next
+            p, q = p.next, q.next
         return p.val
 ```
 

lcci/02.02.Kth Node From End of List/Solution.py

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

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

@@ -8,9 +8,7 @@
 
 <p>现有一个链表 --&nbsp;head =&nbsp;[4,5,1,9]，它可以表示为:</p>
 
-<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>
-
-<p>&nbsp;</p>
+![](./images/237_example.png)
 
 <p><strong>示例 1:</strong></p>
 
@@ -49,19 +47,60 @@
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```python
-
+# Definition for singly-linked list.
+# class ListNode:
+#     def __init__(self, x):
+#         self.val = x
+#         self.next = None
+
+class Solution:
+    def deleteNode(self, node):
+        """
+        :type node: ListNode
+        :rtype: void Do not return anything, modify node in-place instead.
+        """
+        node.val = node.next.val
+        node.next = node.next.next
 ```
 
 ### **Java**
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```java
-
-```
-
-### **...**
+/**
+ * Definition for singly-linked list.
+ * public class ListNode {
+ *     int val;
+ *     ListNode next;
+ *     ListNode(int x) { val = x; }
+ * }
+ */
+class Solution {
+    public void deleteNode(ListNode node) {
+        node.val = node.next.val;
+        node.next = node.next.next;
+    }
+}
 ```
 
+### **JavaScript**
+
+```javascript
+/**
+ * Definition for singly-linked list.
+ * function ListNode(val) {
+ *     this.val = val;
+ *     this.next = null;
+ * }
+ */
+/**
+ * @param {ListNode} node
+ * @return {void} Do not return anything, modify node in-place instead.
+ */
+var deleteNode = function(node) {
+    node.val = node.next.val
+    node.next = node.next.next
+};
 ```
 
 <!-- tabs:end -->

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

@@ -60,15 +60,10 @@
 
 
 <ul>
-
 	<li>The linked list will have at least two elements.</li>
-
 	<li>All of the nodes&#39; values will be unique.</li>
-
 	<li>The given node&nbsp;will not be the tail and it will always be a valid node of the linked list.</li>
-
 	<li>Do not return anything from your function.</li>
-
 </ul>
 
 
@@ -82,18 +77,59 @@
 ### **Python3**
 
 ```python
-
+# Definition for singly-linked list.
+# class ListNode:
+#     def __init__(self, x):
+#         self.val = x
+#         self.next = None
+
+class Solution:
+    def deleteNode(self, node):
+        """
+        :type node: ListNode
+        :rtype: void Do not return anything, modify node in-place instead.
+        """
+        node.val = node.next.val
+        node.next = node.next.next
 ```
 
 ### **Java**
 
 ```java
-
-```
-
-### **...**
+/**
+ * Definition for singly-linked list.
+ * public class ListNode {
+ *     int val;
+ *     ListNode next;
+ *     ListNode(int x) { val = x; }
+ * }
+ */
+class Solution {
+    public void deleteNode(ListNode node) {
+        node.val = node.next.val;
+        node.next = node.next.next;
+    }
+}
 ```
 
+### **JavaScript**
+
+```javascript
+/**
+ * Definition for singly-linked list.
+ * function ListNode(val) {
+ *     this.val = val;
+ *     this.next = null;
+ * }
+ */
+/**
+ * @param {ListNode} node
+ * @return {void} Do not return anything, modify node in-place instead.
+ */
+var deleteNode = function(node) {
+    node.val = node.next.val
+    node.next = node.next.next
+};
 ```
 
 <!-- tabs:end -->

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

@@ -1,14 +1,14 @@
+/**
+ * Definition for singly-linked list.
+ * public class ListNode {
+ *     int val;
+ *     ListNode next;
+ *     ListNode(int x) { val = x; }
+ * }
+ */
 class Solution {
     public void deleteNode(ListNode node) {
-        // 保存下一个结点
-        ListNode tmp = node.next;
-        
-        // 将下个结点的值赋给当前要删除的结点
         node.val = node.next.val;
         node.next = node.next.next;
-        
-        // tmp 置为空，让jvm进行垃圾回收
-        tmp = null;
-        
     }
 }

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

@@ -1,4 +1,15 @@
-const deleteNode = function(node){
-  node.val = node.next.val;
-  node.next = node.next.next;
-}
+/**
+ * Definition for singly-linked list.
+ * function ListNode(val) {
+ *     this.val = val;
+ *     this.next = null;
+ * }
+ */
+/**
+ * @param {ListNode} node
+ * @return {void} Do not return anything, modify node in-place instead.
+ */
+var deleteNode = function(node) {
+  node.val = node.next.val
+  node.next = node.next.next
+};

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

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