25. Reverse Nodes in k-Group
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:
思路:
题目意思是每k个节点翻转一次链表,如果节点不足k个,就不翻转,这是一个简单的实现题,运用递归的思想,找到翻转链表的起始点,然后翻转k个节点。
代码:
java:
/**
* 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 count = 0;
ListNode cur = head;
while (cur != null && count != k) {
cur = cur.next;
count++;
}
if (count == k) {
cur = reverseKGroup(cur, k);
// resever list
while (count-- > 0) {
ListNode tmp = head.next;
head.next = cur;
cur = head;
head = tmp;
}
head = cur;
}
return head;
}
}