Perfected 707-design-linked-list.md.

gazeldx · gazeldx · commit 1c57b9c0898d · 2025-03-15T10:58:05.000+08:00
diff --git a/en/1-1000/349-intersection-of-two-arrays.md b/en/1-1000/349-intersection-of-two-arrays.md
@@ -23,7 +23,7 @@ Each element in the result must be **unique** and you may return the result in *
 
 ## Intuition
 1. Convert one of the arrays to a `set`. The elements are unique in a `set`.
-2. When traversing the other array, if the an element is found to already exist in the `set`, it means that the element belongs to the intersection, and the element should be added to the `results`.
+2. When traversing the other array, if an element is found to already exist in the `set`, it means that the element belongs to the intersection, and the element should be added to the `results`.
 3. The `results` is also of `set` type because duplicate removal is required.
 
 ## Complexity
diff --git a/en/1-1000/707-design-linked-list.md b/en/1-1000/707-design-linked-list.md
@@ -14,12 +14,12 @@ If you want to use the doubly linked list, you will need one more attribute `pre
 
 Implement the `MyLinkedList` class:
 
-* `MyLinkedList()` Initializes the `MyLinkedList` object.
-* `int get(int index)` Get the value of the `index-th` node in the linked list. If the index is invalid, return `-1`.
-* `void addAtHead(int val)` 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.
-* `void addAtTail(int val)` Append a node of value `val` as the last element of the linked list.
-* `void addAtIndex(int index, int val)` Add a node of value `val` before the `index-th` node in the linked list. If `index` equals the length of the linked list, the node will be appended to the end of the linked list. If `index` is greater than the length, the node will **not be inserted**.
-* `void deleteAtIndex(int index)` Delete the `index-th` node in the linked list, if the index is valid.
+- `MyLinkedList()` Initializes the `MyLinkedList` object.
+- `int get(int index)` Get the value of the `index-th` node in the linked list. If the index is invalid, return `-1`.
+- `void addAtHead(int val)` 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.
+- `void addAtTail(int val)` Append a node of value `val` as the last element of the linked list.
+- `void addAtIndex(int index, int val)` Add a node of value `val` before the `index-th` node in the linked list. If `index` equals the length of the linked list, the node will be appended to the end of the linked list. If `index` is greater than the length, the node will **not be inserted**.
+- `void deleteAtIndex(int index)` Delete the `index-th` node in the linked list, if the index is valid.
 
 ### [Example 1]
 **Input**
diff --git a/zh/1-1000/707-design-linked-list.md b/zh/1-1000/707-design-linked-list.md
@@ -5,21 +5,20 @@ LeetCode link: [707. Design Linked List](https://leetcode.com/problems/design-li
 [中文题解](#中文题解)
 
 ## LeetCode problem description
-Design your implementation of the linked list. You can choose to use a singly or doubly linked list.
+你可以选择使用单链表或者双链表，设计并实现自己的链表。
 
-A node in a singly linked list should have two attributes: `val` and `next`. `val` is the value of the current node, and `next` is a pointer/reference to the next node.
+单链表中的节点应该具备两个属性：`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。
 
-If you want to use the doubly linked list, you will need one more attribute `prev` to indicate the previous node in the linked list. Assume all nodes in the linked list are **0-indexed**.
+如果是双向链表，则还需要属性 `prev` 以指示链表中的上一个节点。假设链表中的所有节点下标从 0 开始。
 
+实现 `MyLinkedList` 类：
 
-Implement the `MyLinkedList` class:
-
-* `MyLinkedList()` Initializes the `MyLinkedList` object.
-* `int get(int index)` Get the value of the `index-th` node in the linked list. If the index is invalid, return `-1`.
-* `void addAtHead(int val)` 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.
-* `void addAtTail(int val)` Append a node of value `val` as the last element of the linked list.
-* `void addAtIndex(int index, int val)` Add a node of value `val` before the `index-th` node in the linked list. If `index` equals the length of the linked list, the node will be appended to the end of the linked list. If `index` is greater than the length, the node will **not be inserted**.
-* `void deleteAtIndex(int index)` Delete the `index-th` node in the linked list, if the index is valid.
+- `MyLinkedList()` 初始化 `MyLinkedList` 对象。
+- `int get(int index)` 获取链表中下标为 `index` 的节点的值。如果下标无效，则返回 `-1` 。
+- `void addAtHead(int val)` 将一个值为 `val` 的节点插入到链表中第一个元素之前。在插入完成后，新节点会成为链表的第一个节点。
+- `void addAtTail(int val)` 将一个值为 `val` 的节点追加到链表中作为链表的最后一个元素。
+- `void addAtIndex(int index, int val)` 将一个值为 `val` 的节点插入到链表中下标为 `index` 的节点之前。如果 `index` 等于链表的长度，那么该节点会被追加到链表的末尾。如果 index 比长度更大，该节点将 不会插入 到链表中。
+- `void deleteAtIndex(int index)` 如果下标有效，则删除链表中下标为 `index` 的节点。
 
 ### [Example 1]
 **Input**
@@ -38,16 +37,16 @@ Implement the `MyLinkedList` class:
 MyLinkedList myLinkedList = new MyLinkedList();
 myLinkedList.addAtHead(1);
 myLinkedList.addAtTail(3);
-myLinkedList.addAtIndex(1, 2);    // linked list becomes 1->2->3
-myLinkedList.get(1);              // return 2
-myLinkedList.deleteAtIndex(1);    // now the linked list is 1->3
-myLinkedList.get(1);              // return 3
+myLinkedList.addAtIndex(1, 2);    // 链表变为 1->2->3
+myLinkedList.get(1);              // 返回 2
+myLinkedList.deleteAtIndex(1);    // 现在，链表变为 1->3
+myLinkedList.get(1);              // 返回 3
 ```
 
 ### [Constraints]
 - `0 <= index, val <= 1000`
-- Please do not use the built-in LinkedList library.
-- At most `2000` calls will be made to `get`, `addAtHead`, `addAtTail`, `addAtIndex` and `deleteAtIndex`.
+- 请不要使用内置的 LinkedList 库。
+- 调用 `get`、`addAtHead`、`addAtTail`、`addAtIndex` 和 `deleteAtIndex` 的次数不超过 `2000` 。
 
 ## Intuition
 Before solving this problem, it is recommended to solve the simple problem [19. Remove Nth Node From End of List](19-remove-nth-node-from-end-of-list.md) first.
