feat: update 92-reverse-between

YanceyOfficial · Nov 17, 2023 · bee3eb5 · bee3eb5
1 parent f9b14a0
commit bee3eb5
Showing 1 changed file with 5 additions and 64 deletions.
diff --git a/leetcode-docs/medium/92-reverse-between.mdx b/leetcode-docs/medium/92-reverse-between.mdx
@@ -38,44 +38,7 @@ keywords:
 
 ## 题解
 
-import Tabs from '@theme/Tabs'
-import TabItem from '@theme/TabItem'
-
-<Tabs>
-  <TabItem value="JavaScript - 递归" label="JavaScript - 递归">
-
-在做这道题之前, 先整个前菜, 也就是**反转链表前 N 个节点**, 这个题跟 [206. 反转链表](/leetcode/easy/206-reverse-list) 差不多, 因为是反转前 N 个, 因此得把第 N + 1 个节点记录下来, 以拼接在反转子链表之后.
-
 ```ts
-let successor = null // 后驱节点
-
-/**
- * Definition for singly-linked list.
- * function ListNode(val) {
- *     this.val = val;
- *     this.next = null;
- * }
- */
-/**
- * @param {ListNode} head
- * @param {number} n
- * @return {ListNode}
- */
-var reverseN = function (head, n) {
-  if (n === 1) {
-    // 记录第 n + 1 个节点
-    successor = head.next
-    return head
-  }
-  // 以 head.next 为起点, 需要反转前 n - 1 个节点
-  const last = reverseN(head.next, n - 1)
-
-  head.next.next = head
-  // 让反转之后的 head 节点和后面的节点连起来
-  head.next = successor
-  return last
-}
-
 /**
  * Definition for singly-linked list.
  * function ListNode(val) {
@@ -89,48 +52,26 @@ var reverseN = function (head, n) {
  * @param {number} n
  * @return {ListNode}
  */
-var reverseBetween = function (head, m, n) {
-  // 如果 m === 1, 就相当于反转链表开头的 n 个元素
-  if (m === 1) {
-    return reverseN(head, n)
-  }
-  // 前进到反转的起点触发 base case
-  head.next = reverseBetween(head.next, m - 1, n - 1)
-  return head
-}
-```
-
-</TabItem>
-<TabItem value="JavaScript - 迭代" label="JavaScript - 迭代" default>
-
-- 先创建一个 `dummyHead`, 它的 `next` 指向 `head`;
-- 分别创建两个节点 `g(guard 节点)` 和 `p(point 节点)`. 其中前者指向 `dummyHead`, 后者指向 `dummyHead.next`;
-- 以 m = 2, n = 4 为例, 将 g 移动到**第一个要反转的节点的前面**, 即节点 1; 将 p 移动到**第一个要反转的节点的位置上**, 即节点 2;
-- 接下来是使用"头插法", 将 `p` 后面的去掉, 然后添加到 `g` 的后面.
-
-```ts
 var reverseBetween = function (head, m, n) {
   const dummyHead = new ListNode(0)
   dummyHead.next = head
 
   let g = dummyHead,
     p = dummyHead.next
 
-  for (let step = 0; step < m - 1; step++) {
+  // 先找到需要反转的起点
+  for (let i = 0; i < m - 1; i++) {
     g = g.next
     p = p.next
   }
 
   for (let i = 0; i < n - m; i++) {
     const removed = p.next
-    p.next = p.next.next // 将要反转的节点去掉, 比如去掉 3, 此时 p 成了 2 -> 4 -> 5
-    removed.next = g.next // removed 此时为 3 -> 4 -> 5, 将 g.next 移过来, 成了 3 -> 2 -> 4 -> 5
-    g.next = removed // 将新的 remove 放到 g.next 即可
+    p.next = p.next.next
+    removed.next = g.next
+    g.next = removed
   }
 
   return dummyHead.next
 }
 ```
-
-</TabItem>
-</Tabs>