LeetCode Reverse Nodes in k-Group(单链表连续分段反转)

题意：给出一个单链表及反转元素的个数，求反转后的单链表

思路：首先确定链表总结点数，根据反转元素的个数，可以知道要反转多少次。第一次反转的尾为新的单链表的头，而后序反转后的头变成前一次反转后发尾部元素的后继结点

具体代码如下：

public class Solution
{
public ListNode reverseKGroup(ListNode head, int k)
{
if (null == head || 0 == k || 1 == k) return head;

int n = 0;
ListNode cur = head;

while (cur != null)
{
n++;
cur = cur.next;
}

cur = head;

int cnt = n / k, remainder = n % k;

ListNode ans = null;
ListNode tail = null;
for (int i = 0; i < cnt; i++)
{
ListNode pre = null;
ListNode tmp = cur;
int c = 0;
while (c++ < k)
{
ListNode next = cur.next;
cur.next = pre;
pre = cur;
cur = next;
}

if (i != 0)
{
tail.next = pre;
tail = tmp;
}
else
{
tail = tmp;
ans = pre;
}
}

if (remainder != 0)
{
if (tail != null) tail.next = cur;
}

if (null == ans)  ans = head;

return ans;
}