题解 | #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
https://www.nowcoder.com/practice/b49c3dc907814e9bbfa8437c251b028e
import java.util.*;
/*
* public class ListNode {
* int val;
* ListNode next = null;
* public ListNode(int val) {
* this.val = val;
* }
* }
*/
public class Solution {
/**
* 代码中的类名、方法名、参数名已经指定,请勿修改,直接返回方法规定的值即可
*
*
* @param head ListNode类
* @param k int整型
* @return ListNode类
*/
public ListNode reverseKGroup (ListNode head, int k) {
// write code here
ListNode dummy = new ListNode(-1);
dummy.next = head;
ListNode pre = dummy;
ListNode end = dummy;
while(end != null){
for(int i = 0;i<k && end != null;i++) end = end.next;
if(end == null) break;
ListNode next = end.next;
ListNode start = pre.next;
end.next = null;
pre.next = reverse(start);
start.next = next;
pre = start;
end = start;
}
return dummy.next;
}
public ListNode reverse(ListNode head){
ListNode pre = null;
ListNode cur = head;
while(cur != null){
ListNode next = cur.next;
cur.next = pre;
pre = cur;
cur = next;
}
return pre;
}
}
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