题解 | #BM3 链表中的节点每k个一组翻转#
链表中的节点每k个一组翻转
http://www.nowcoder.com/practice/b49c3dc907814e9bbfa8437c251b028e
import java.util.*;
/*
* public class ListNode {
* int val;
* ListNode next = null;
* }
*/
public class Solution {
/**
*
* @param head ListNode类
* @param k int整型
* @return ListNode类
*/
public ListNode reverseKGroup (ListNode head, int k) {
ListNode top = new ListNode(0);
ListNode pre=top;
ListNode tail = top;
if(k<2){
return head;
}
int i=1;
while( head!=null ){
if(i==k+1){
pre=tail;
i=1;
}
if(i==1){
tail = head;
}
ListNode temp = head.next;
head.next = pre.next;
pre.next = head;
head = temp;
i++;
}
if(i<=k&&i>2){
head = pre.next;
pre.next=null;
while(head!=null){
ListNode temp = head.next;
head.next = pre.next;
pre.next = head;
head = temp;
}
}
return top.next;
}
}
解体思路：不管是不是满足k个，直接反转。最后判断最后一个是不是k个，如果不满足k，则再反转回去即可。
