题目:

给你链表的头节点 head ,每 k 个节点一组进行翻转,请你返回修改后的链表。

k 是一个正整数,它的值小于或等于链表的长度。如果节点总数不是 k 的整数倍,那么请将最后剩余的节点保持原有顺序。

你不能只是单纯的改变节点内部的值,而是需要实际进行节点交换。

 

示例 1:

输入:head = [1,2,3,4,5], k = 2

输出:[2,1,4,3,5]

示例 2:

输入:head = [1,2,3,4,5], k = 3

输出:[3,2,1,4,5]

代码实现:

class Solution {
public ListNode reverseKGroup(ListNode head, int k) {
if(head == null || head.next == null) return head;

ListNode dummyHead = new ListNode(0);
dummyHead.next = head;
ListNode cur = head;
int count = 0;
while(cur != null)
{
cur = cur.next;
count++;
}
int timesToFlip = count / k;
cur= head;
ListNode before = dummyHead;
ListNode after = null;
for(int i = 0; i < timesToFlip; i++)
{
int numNodes = k;
ListNode temp = null, pre = null;

//Reverse partial Linked-list
while(numNodes != 0)
{
numNodes--;

temp = cur.next;
cur.next = pre;
pre = cur;
cur = temp;
after = cur;
}
before.next = pre;
while(pre.next != null)
pre = pre.next;
pre.next = after;
before = pre;
cur = before.next;
}

return dummyHead.next;
}
}