标签：nbsp this linked ant tip odi for pre out
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:
M1: recursive
cur指向head，用cnt计数，先走k步。当走了k步时（即cnt == k），递归，然后反转这一部分的链表
time: O(n), space: O(n)
/** * 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) { if(head == null || head.next == null) { return head; } int cnt = 0; ListNode cur = head; while(cur != null && cnt < k) { cur = cur.next; cnt++; } if(cnt == k) { cur = reverseKGroup(cur, k); while(cnt-- > 0) { ListNode next = head.next; head.next = cur; cur = head; head = next; } head = cur; } return head; } }
25. Reverse Nodes in k-Group - Hard
标签：nbsp this linked ant tip odi for pre out
原文地址：https://www.cnblogs.com/fatttcat/p/10180540.html
