leetcode -- 25. Reverse Nodes in k-Group

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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;}
}

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