题目描述

题目难度：Hard
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:

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

Note:

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

AC代码

借鉴自 leetcode 上面大神的代码

/** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { val = x; } * } */
class Solution {
    public ListNode reverseKGroup(ListNode head, int k) {
        ListNode dummy = new ListNode(0);
        dummy.next = head;
        
        ListNode begin = dummy;
        ListNode end = begin.next;
        
        int i = 0;
        while(end != null) {
            i++;
            if(i % k == 0) {
                begin = reverse(begin, end.next);
                end = begin.next;
            }
            else {
                end = end.next;
            }
        } 
        
        return dummy.next;
    }
    
    //愣是没搞懂，得画个图才能明白
    private ListNode reverse(ListNode begin, ListNode end) {
        ListNode curr = begin.next, prev = begin;
        ListNode first = curr, next;
        
        while(curr != end) {
            next = curr.next;
            curr.next = prev;
            prev = curr;
            curr = next;
        }
        
        begin.next = prev;
        first.next = curr;
        
        return first;
    }
}