LeetCode（59）反转链表 II【链表】【中等】

news2025/12/20 19:54:53

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1.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 。

示例 1：

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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？

2.答案

/**
 * 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 static ListNode reverseBetween(ListNode head, int left, int right) {
        if (left == right) {
            return head;
        }
        ListNode node = head;
        int index = 1;
        ListNode beforeLeftNode = null;
        ListNode leftNode = null;
        ListNode rightNode = null;
        ListNode afterRightNode = null;
        ListNode beforeNode = null;
        while (node != null) {
            if (index == left) {
                beforeLeftNode = beforeNode;
                leftNode = node;
                beforeNode = node;
                node = node.next;
            } else if (index > left && index < right) {
                ListNode afterNode = node.next;
                node.next = beforeNode;
                beforeNode = node;
                node = afterNode;
            } else if (index == right) {
                afterRightNode = node.next;
                rightNode = node;
                rightNode.next = beforeNode;
                break;
            } else {
                beforeNode = node;
                node = node.next;
            }
            index++;
        }
        if (beforeLeftNode != null) {
            beforeLeftNode.next = rightNode;
        } else {
            head = rightNode;
        }
        assert leftNode != null;
        leftNode.next = afterRightNode;
        return head;
    }
}