📝 add lc 92

DreamCats · DreamCats · commit 42bc09feb533 · 2021-04-13T13:25:35.000+08:00
diff --git a/Java/alg/README.md b/Java/alg/README.md
@@ -130,7 +130,8 @@
 - [82.删除排序链表中的重复元素2](lc/82.删除排序链表中的重复元素2.md)
 - [86.分隔链表](lc/86.分隔链表.md)
 - [90.子集2](lc/90.子集2.md)
-- [91.解码方法](lc/91. 解码方法.md)
+- [91.解码方法](lc/91.解码方法.md)
+- [92.反转链表2](92.反转链表2.md)
 
 ## 笔试
 
diff --git a/Java/alg/lc/92.反转链表2.md b/Java/alg/lc/92.反转链表2.md
@@ -0,0 +1,96 @@
+# 92. 反转链表 II
+
+[url](https://leetcode-cn.com/problems/reverse-linked-list-ii/)
+
+## 题目
+
+给你单链表的头指针 head 和两个整数 left 和 right ，其中 left <= right 。请你反转从位置 left 到位置 right 的链表节点，返回 反转后的链表 。
+
+[](https://assets.leetcode.com/uploads/2021/02/19/rev2ex2.jpg)
+
+
+```
+输入：head = [1,2,3,4,5], left = 2, right = 4
+输出：[1,4,3,2,5]
+输入：head = [5], left = 1, right = 1
+输出：[5]
+```
+
+## 方法
+
+
+## code
+
+### js
+
+```js
+let reverseBetween = (head, m, n) => {
+    let p = head;
+    let idx = 1;
+    let stack = [];
+    while (p != null) {
+        if (idx >= m && idx <= n)
+            stack.push(p.val)
+        idx++
+        p = p.next;
+    }
+    idx = 1;
+    p = head;
+    while (p != null) {
+        if (idx >= m && idx <= n)
+            p.val = stack.pop();
+        idx++;
+        p = p.next;
+    }
+    return head;
+}
+```
+
+### go
+
+```go
+func reverseBetween(head *ListNode, m int, n int) *ListNode {
+    dummy := &ListNode{}
+    dummy.Next = head
+    pre, cur, next := dummy, &ListNode{}, &ListNode{}
+    for i := 1; i < m; i++ {
+        pre = pre.Next
+    }
+    cur = pre.Next
+    for i := m; i < n; i++ {
+        next = cur.Next
+        cur.Next = next.Next
+        next.Next = pre.Next
+        pre.Next = next
+    }
+    return dummy.Next
+}
+```
+
+### java
+
+```java
+class Solution {
+    public ListNode reverseBetween(ListNode head, int m, int n) {
+        ListNode p = head;
+        int idx = 1;
+        Stack<Integer> stack = new Stack<>();
+        while (p != null) {
+            if (idx >= m && idx <= n)
+                stack.push(p.val);
+            idx++;
+            p = p.next;
+        }
+        idx = 1;
+        p = head;
+        while (p != null) {
+            if (idx >= m && idx <= n)
+                p.val = stack.pop();
+            idx++;
+            p = p.next;
+        }
+        return head;
+    }
+}
+```
+
