92:反转链表 II

src/main/java/leetcode/editor/cn/ReverseLinkedListIi.java

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+//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 。
+// 
+//
+// 示例 1： 
+//
+// 
+//输入：head = [1,2,3,4,5], left = 2, right = 4
+//输出：[1,4,3,2,5]
+// 
+//
+// 示例 2： 
+//
+// 
+//输入：head = [5], left = 1, right = 1
+//输出：[5]
+// 
+//
+// 
+//
+// 提示： 
+//
+// 
+// 链表中节点数目为 n 
+// 1 <= n <= 500 
+// -500 <= Node.val <= 500 
+// 1 <= left <= right <= n 
+// 
+//
+// 
+//
+// 进阶： 你可以使用一趟扫描完成反转吗？ 
+// Related Topics 链表 
+// 👍 916 👎 0
+
+package leetcode.editor.cn;
+
+import com.code.leet.entiy.ListNode;
+
+import java.util.ArrayList;
+import java.util.List;
+
+//92:反转链表 II
+public class ReverseLinkedListIi{
+    public static void main(String[] args) {
+        //测试代码
+        Solution solution = new ReverseLinkedListIi().new Solution();
+    }
+    //力扣代码
+    //leetcode submit region begin(Prohibit modification and deletion)
+/**
+ * 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 reverseBetween(ListNode head, int m, int n) {
+        if (n == 1) {
+            return head;
+        }
+        List<ListNode> list = new ArrayList<>();
+        int num = 1;
+        ListNode before = new ListNode(0);
+        ListNode after = null;
+        ListNode temp = head;
+        while (temp != null) {
+            if (num == m - 1) {
+                before = temp;
+            }
+            if (num >= m && num <= n) {
+                list.add(temp);
+            }
+            if (num == n) {
+                after = temp.next;
+                break;
+            }
+            temp = temp.next;
+            num++;
+        }
+        ListNode newHead = before;
+        int size = list.size();
+        for (int i = size - 1; i >= 0; i--) {
+            before.next = list.get(i);
+            before = before.next;
+        }
+        before.next = after;
+        if (m == 1) {
+            head = newHead.next;
+        }
+        return head;
+    }
+}
+//leetcode submit region end(Prohibit modification and deletion)
+
+}

src/main/java/leetcode/editor/cn/ReverseLinkedListIi.md

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+给你单链表的头指针 <code>head</code> 和两个整数 <code>left</code> 和 <code>right</code> ，其中 <code>left <= right</code> 。请你反转从位置 <code>left</code> 到位置 <code>right</code> 的链表节点，返回 <strong>反转后的链表</strong> 。
+<p> </p>
+
+<p><strong>示例 1：</strong></p>
+<img alt="" src="https://assets.leetcode.com/uploads/2021/02/19/rev2ex2.jpg" style="width: 542px; height: 222px;" />
+<pre>
+<strong>输入：</strong>head = [1,2,3,4,5], left = 2, right = 4
+<strong>输出：</strong>[1,4,3,2,5]
+</pre>
+
+<p><strong>示例 2：</strong></p>
+
+<pre>
+<strong>输入：</strong>head = [5], left = 1, right = 1
+<strong>输出：</strong>[5]
+</pre>
+
+<p> </p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li>链表中节点数目为 <code>n</code></li>
+	<li><code>1 <= n <= 500</code></li>
+	<li><code>-500 <= Node.val <= 500</code></li>
+	<li><code>1 <= left <= right <= n</code></li>
+</ul>
+
+<p> </p>
+
+<p><strong>进阶：</strong> 你可以使用一趟扫描完成反转吗？</p>
+<div><div>Related Topics</div><div><li>链表</li></div></div>\n<div><li>👍 916</li><li>👎 0</li></div>