题解 | #牛群的重新排列#

import java.util.*;

/*
 * public class ListNode {
 *   int val;
 *   ListNode next = null;
 *   public ListNode(int val) {
 *     this.val = val;
 *   }
 * }
 */

public class Solution {
    /**
    e:\JavaProjects\myTest\src\Solution.java     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     *
     * @param head ListNode类
     * @param left int整型
     * @param right int整型
     * @return ListNode类
     */

    public ListNode reverseBetween (ListNode head, int left, int right) {
        // write code here
        List<Integer> list = new LinkedList<>();
        while (head != null) {
            list.add(head.val);
            head = head.next;
        }
        ListNode node = new ListNode(0);
        ListNode result = node;

        LinkedList<Integer> reverse = new LinkedList<>();
        for (int i = left - 1; i < right; i++) {
            reverse.add(list.get(i));
        }
        for (int i = 0; i < left - 1; i++) {
            node.next = new ListNode(list.get(i));
            node = node.next;
        }
        for (int i = reverse.size() - 1; i >= 0; i--) {
            node.next = new ListNode(reverse.get(i));
            node = node.next;
        }
        for (int i = right; i < list.size(); i++) {
            node.next = new ListNode(list.get(i));
            node = node.next;
        }
        return result.next;
    }
}

本题考察的知识点是链表，所用编程语言是java。

我采用LinkedList将链表节点值按顺序加入当中，我们将left到right中的链表节点值加入到一个新的LinkedList中，然后根据题目要求重建链表