算法题Reverse Nodes in k-Group

最新推荐文章于 2022-05-23 04:03:01 发布

Jack_Weng

最新推荐文章于 2022-05-23 04:03:01 发布

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本文链接：https://blog.youkuaiyun.com/wj310298/article/details/49151777

本文介绍了一种链表K节点反转算法，该算法能够将链表中的节点按K节点进行反转，对于剩余节点则保持不变。算法仅使用常量内存，并在不改变节点值的情况下修改节点本身。

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题目如下

Given a linked list, reverse the nodes of a linked list k at a time and return its modified list.

If the number of nodes is not a multiple of k then left-out nodes in the end should remain as it is.

You may not alter the values in the nodes, only nodes itself may be changed.

Only constant memory is allowed.

For example,
Given this linked list: 1->2->3->4->5

For k = 2, you should return: 2->1->4->3->5

For k = 3, you should return: 3->2->1->4->5

解决源码如下

import java.util.Queue;
import java.util.LinkedList;

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode reverseKGroup(ListNode head, int k) {
        Queue<Integer> q = new LinkedList<Integer> ();
        reverseKGroup(head, q, k, k);
        return head;
    }
    
    boolean stop = false;
    
    ListNode reverseKGroup(ListNode node, Queue<Integer> q, int k, int i) {
        if (node == null) return null;
        q.add(node.val);
        if (i != 1 && node.next == null) {
            stop = true;
            return null;
        }
        if (i == 1) {
            node.val = q.poll();
            return node.next;
        } else if (k == i) {
            ListNode next = reverseKGroup(node.next, q, k, i - 1);
            if (stop) return null;
            node.val = q.poll();
            return reverseKGroup(next, q, k, k);
        } else {
            ListNode tmp = reverseKGroup(node.next, q, k, i - 1);
            if (stop) return null;
            node.val = q.poll();
            return tmp;
        }
    }
}