feat: add solutions to lc problem: No.1019. Next Greater Node in Linked List

Author: yanglbme

README.md

@@ -14,7 +14,7 @@
 
 ## 介绍
 
-本项目包含 [LeetCode](https://leetcode-cn.com/problemset/all/)、[《剑指 Offer（第 2 版）》](https://leetcode-cn.com/problemset/lcof/)、[《程序员面试金典（第 6 版）》](https://leetcode-cn.com/problemset/lcci/)等题目的相关题解。所有题解均由多种编程语言实现，包括但不限于：Java、Python、C++、JavaScript、C#、Go，日常更新。欢迎 Star 关注本项目「[GitHub](https://github.com/doocs/leetcode) / [Gitee](https://gitee.com/doocs/leetcode)」，获取项目最新动态。
+本项目包含 [LeetCode](https://leetcode-cn.com/problemset/all/)、[《剑指 Offer（第 2 版）》](https://leetcode-cn.com/problemset/lcof/)、[《程序员面试金典（第 6 版）》](https://leetcode-cn.com/problemset/lcci/)等题目的相关题解。所有题解均由多种编程语言实现，包括但不限于：Java、Python、C++、JavaScript、C#、Go，日常更新。欢迎 Star 🌟 关注本项目「[GitHub](https://github.com/doocs/leetcode) / [Gitee](https://gitee.com/doocs/leetcode)」，获取项目最新动态。
 
 [English Version](/README_EN.md)
 

solution/0800-0899/0876.Middle of the Linked List/README.md

@@ -38,27 +38,57 @@ ans.val = 3, ans.next.val = 4, ans.next.next.val = 5, 以及 ans.next.next.next
 	<li>给定链表的结点数介于 <code>1</code> 和 <code>100</code> 之间。</li>
 </ul>
 
-
 ## 解法
 
 <!-- 这里可写通用的实现逻辑 -->
 
+“快慢指针”实现。
+
 <!-- tabs:start -->
 
 ### **Python3**
 
 <!-- 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, val=0, next=None):
+#         self.val = val
+#         self.next = next
+class Solution:
+    def middleNode(self, head: ListNode) -> ListNode:
+        slow = fast = head
+        while fast and fast.next:
+            slow, fast = slow.next, fast.next.next
+        return slow
 ```
 
 ### **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() {}
+ *     ListNode(int val) { this.val = val; }
+ *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
+ * }
+ */
+class Solution {
+    public ListNode middleNode(ListNode head) {
+        ListNode slow = head, fast = head;
+        while (fast != null && fast.next != null) {
+            slow = slow.next;
+            fast = fast.next.next;
+        }
+        return slow;
+    }
+}
 ```
 
 ### **...**

solution/0800-0899/0876.Middle of the Linked List/README_EN.md

@@ -6,22 +6,14 @@
 
 <p>Given a non-empty, singly&nbsp;linked list with head node <code>head</code>, return&nbsp;a&nbsp;middle node of linked list.</p>
 
-
-
 <p>If there are two middle nodes, return the second middle node.</p>
 
-
-
 <p>&nbsp;</p>
 
-
-
 <div>
 
 <p><strong>Example 1:</strong></p>
 
-
-
 <pre>
 
 <strong>Input: </strong><span id="example-input-1-1">[1,2,3,4,5]</span>
@@ -36,14 +28,10 @@ ans.val = 3, ans.next.val = 4, ans.next.next.val = 5, and ans.next.next.next = N
 
 </pre>
 
-
-
 <div>
 
 <p><strong>Example 2:</strong></p>
 
-
-
 <pre>
 
 <strong>Input: </strong><span id="example-input-2-1">[1,2,3,4,5,6]</span>
@@ -54,16 +42,10 @@ Since the list has two middle nodes with values 3 and 4, we return the second on
 
 </pre>
 
-
-
 <p>&nbsp;</p>
 
-
-
 <p><strong>Note:</strong></p>
 
-
-
 <ul>
 	<li>The number of nodes in the given list will be between <code>1</code>&nbsp;and <code>100</code>.</li>
 </ul>
@@ -72,22 +54,49 @@ Since the list has two middle nodes with values 3 and 4, we return the second on
 
 </div>
 
-
-
 ## Solutions
 
 <!-- tabs:start -->
 
 ### **Python3**
 
 ```python
-
+# Definition for singly-linked list.
+# class ListNode:
+#     def __init__(self, val=0, next=None):
+#         self.val = val
+#         self.next = next
+class Solution:
+    def middleNode(self, head: ListNode) -> ListNode:
+        slow = fast = head
+        while fast and fast.next:
+            slow, fast = slow.next, fast.next.next
+        return slow
 ```
 
 ### **Java**
 
 ```java
-
+/**
+ * Definition for singly-linked list.
+ * public class ListNode {
+ *     int val;
+ *     ListNode next;
+ *     ListNode() {}
+ *     ListNode(int val) { this.val = val; }
+ *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
+ * }
+ */
+class Solution {
+    public ListNode middleNode(ListNode head) {
+        ListNode slow = head, fast = head;
+        while (fast != null && fast.next != null) {
+            slow = slow.next;
+            fast = fast.next.next;
+        }
+        return slow;
+    }
+}
 ```
 
 ### **...**

solution/0800-0899/0876.Middle of the Linked List/Solution.java

@@ -1,8 +1,20 @@
-public ListNode middleNode(ListNode head) {
-    ListNode low = head, first = head;
-    while (first != null && first.next != null) {
-        low = low.next;
-        first = first.next.next;
+/**
+ * Definition for singly-linked list.
+ * public class ListNode {
+ *     int val;
+ *     ListNode next;
+ *     ListNode() {}
+ *     ListNode(int val) { this.val = val; }
+ *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
+ * }
+ */
+class Solution {
+    public ListNode middleNode(ListNode head) {
+        ListNode slow = head, fast = head;
+        while (fast != null && fast.next != null) {
+            slow = slow.next;
+            fast = fast.next.next;
+        }
+        return slow;
     }
-    return low;
 }

solution/0800-0899/0876.Middle of the Linked List/Solution.py

@@ -1,23 +1,11 @@
 # Definition for singly-linked list.
 # class ListNode:
-#     def __init__(self, x):
-#         self.val = x
-#         self.next = None
-
+#     def __init__(self, val=0, next=None):
+#         self.val = val
+#         self.next = next
 class Solution:
-    def middleNode(self, head):
-        """
-        :type head: ListNode
-        :rtype: ListNode
-        """
-        if not head:
-            return None
-        if not head.next:
-            return head
-        fast=head
-        slow=head
-        while fast.next:
-            fast=fast.next.next
-            slow=slow.next
-            if not fast or not fast.next:
-                return slow
+    def middleNode(self, head: ListNode) -> ListNode:
+        slow = fast = head
+        while fast and fast.next:
+            slow, fast = slow.next, fast.next.next
+        return slow

solution/1000-1099/1019.Next Greater Node In Linked List/README.md

@@ -43,27 +43,71 @@
 	<li>给定列表的长度在 <code>[0, 10000]</code>&nbsp;范围内</li>
 </ol>
 
-
 ## 解法
 
 <!-- 这里可写通用的实现逻辑 -->
 
+“单调栈”实现。
+
 <!-- tabs:start -->
 
 ### **Python3**
 
 <!-- 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 nextLargerNodes(self, head: ListNode) -> List[int]:
+        nums = []
+        while head:
+            nums.append(head.val)
+            head = head.next
+        s = []
+        larger = [0] * len(nums)
+        for i, num in enumerate(nums):
+            while s and nums[s[-1]] < num:
+                larger[s.pop()] = num
+            s.append(i)
+        return larger
 ```
 
 ### **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 int[] nextLargerNodes(ListNode head) {
+        List<Integer> nums = new ArrayList<>();
+        while (head != null) {
+            nums.add(head.val);
+            head = head.next;
+        }
+        Deque<Integer> s = new ArrayDeque<>();
+        int[] larger = new int[nums.size()];
+        for (int i = 0; i < nums.size(); ++i) {
+            while (!s.isEmpty() && nums.get(s.peek()) < nums.get(i)) {
+                larger[s.pop()] = nums.get(i);
+            }
+            s.push(i);
+        }
+        return larger;
+    }
+}
 ```
 
 ### **...**