LeetCode 题解-25. Reverse Nodes in k-Group

对Leetcode 25. Reverse Nodes in k-Group 题目的求解。

Problem 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.

Example#

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

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

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

Note#

Only constant extra memory is allowed.
You may not alter the values in the list’s nodes, only nodes itself may be changed.

Analysis#

这道题和206题一样，也是一道链表题目，核心是指针的交换操作。关键点总结如下

在链表最前端设置zero节点。
每次选中k个节点，不停地将cur的next节点挪到start的后面

事实上，代码还可以更简练，通过recursive的方法。事实上，代码还可以更简练，通过recursive的方法。

Solution#

1
class Solution {
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    public static ListNode reverseKGroup(ListNode head, int k) {
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        ListNode zero = new ListNode(0);
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        zero.next = head;
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        ListNode start = zero;
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        ListNode tail = zero;
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        while (true){
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            for (int i = 0; i < k ; ++i){
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                tail = tail.next;
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                if (tail == null) break;
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            }
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            if (tail == null) break;
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            start = reverseK(start, k);
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            tail = start;
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        }
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        return zero.next;
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    }
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    public static ListNode reverseK(ListNode start, int k){
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        ListNode cur = start.next;
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        for (int i = 0; i < k - 1; ++i){
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            ListNode tmp = cur.next.next;
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            cur.next.next = start.next;
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            start.next = cur.next;
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            cur.next =  tmp;
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        }
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        return cur;
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    }
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}