LeetCode Reverse Linked List II

最新推荐文章于 2025-09-11 19:19:55 发布

原创最新推荐文章于 2025-09-11 19:19:55 发布 · 251 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#leetcode #java #LinkedList

算法/oj 专栏收录该内容

253 篇文章

订阅专栏

本文介绍了如何在不使用额外内存的情况下，通过一次遍历链表，实现从指定位置m到n之间的节点进行反转。通过实例演示了具体操作步骤，并解释了相关注意事项。

Description:

Reverse a linked list from position m to n. Do it in-place and in one-pass.

For example:
Given 1->2->3->4->5->NULL, m = 2 and n = 4,

return 1->4->3->2->5->NULL.

Note:
Given m, n satisfy the following condition:
1 ≤ m ≤ n ≤ length of list.

Solution:

只需要按照题目给出的模拟一下，链表的基本操作。注意+1 -1的区间操作。

import java.util.*;

public class Solution {
	public ListNode reverseBetween(ListNode head, int m, int n) {
		if (m == n)
			return head;

		ListNode pre_node_m, next_node_n;
		ListNode neoHead, neoTail, current, next;
		if (m == 1) {
			next_node_n = getNode(head, n + 1);
			neoTail = neoHead = null;
			current = head;
			for (int i = m; i <= n; i++) {
				next = current.next;
				if (neoHead == null) {
					neoHead = neoTail = current;
					current.next = null;
				} else {
					current.next = neoHead;
					neoHead = current;
				}
				current = next;
			}
			head = neoHead;
			neoTail.next = next_node_n;
		} else {
			pre_node_m = getNode(head, m - 1);
			next_node_n = getNode(head, n + 1);

			current = pre_node_m.next;
			neoTail = neoHead = null;
			for (int i = m; i <= n; i++) {
				next = current.next;
				if (neoHead == null) {
					neoHead = neoTail = current;
					current.next = null;
				} else {
					current.next = neoHead;
					neoHead = current;
				}
				current = next;
			}

			pre_node_m.next = neoHead;
			neoTail.next = next_node_n;
		}

		return head;
	}

	ListNode getNode(ListNode head, int n) {
		ListNode temp = head;
		n--;
		while (n > 0) {
			temp = temp.next;
			n--;
		}
		return temp;
	}
}