算法第三周作业03

Description

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

k is a positive integer and is less than or equal to the length of the linked 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

Solution

1. 对于第一组[0,k]，可以从0开始读节点node，另外在一个新链表head里面写，node.next = head, head = node，从而实现倒序；

2. 使用递归方式对接下来的分组[k+1, 2k],[2k+1, 3k]...进行处理；

3. 首先需要读取输入链表的长度，然后判断需要倒序的组数；

Code

public ListNode reverseKGroup(ListNode head, int k) {ListNode oldHead = head;int count = 0;// 获取输入链表的长度，得到需要递归的次数while(head != null){head = head.next;count++;}return recursion(oldHead, k, count/k);}public ListNode recursion(ListNode head, int k, int times){if(head != null && times != 0){ListNode oldHead = head;ListNode node1 = head;ListNode node2 = head.next;int i = 0;head = null;// 读取输入链表// 倒序写入输出链表while (i++ < k && node1 != null) {node1.next = head;head = node1;node1 = node2;if(node1 != null)node2 = node2.next;}// 递归操作接下来的节点oldHead.next = recursion(node1, k, times-1);}return head;}