LeetCode Reverse Nodes in k-Group

Description:

Given a linked list, reverse the nodes of a linked list k at a time and return its modified list.

If the number of nodes is not a multiple of k then left-out nodes in the end should remain as it is.

You may not alter the values in the nodes, only nodes itself may be changed.

Only constant memory is allowed.

For 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

Solution:

一开始还在想有没有什么复杂的优化，最后发现单纯写出来就很好了。

我的写法是最naive的写法，似乎有更好的，等我刷完再慢慢更新哈~~~

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

int i = 0;

ListNode neoHead = head;
for (i = 1; i < k; i++) {
if (neoHead.next == null)
break;
neoHead = neoHead.next;
}
if (i < k)
return head;

ListNode lastTail = null, currentHead, currentTail, nextHead;
ListNode temp, pre, cur;

while (true) {

if (lastTail == null)
currentHead = head;
else
currentHead = lastTail.next;

currentTail = currentHead;
for (i = 1; i < k; i++) {
if (currentTail.next == null)
break;
currentTail = currentTail.next;
}
if (i < k)
return neoHead;

nextHead = currentTail.next;
currentTail.next = null;

if (lastTail == null)
lastTail = currentHead;
else {
lastTail.next = currentTail;
lastTail = currentHead;
}

temp = currentHead.next;
pre = currentHead;
cur = currentHead;
while (temp != null) {
cur = temp;
temp = temp.next;
cur.next = pre;
pre = cur;
}

currentHead.next = nextHead;

if (nextHead == null)
break;
}
return neoHead;
}
}