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
解题思路：
和k=2情况差不多，考虑到后进先出，可以用Stack实现，JAVA代码如下：
public ListNode reverseKGroup(ListNode head, int k) {
ListNode result = new ListNode(0), index = result,temp=head;
if (head == null)
return null;
Stack<Integer> stack =new Stack<Integer>();
while (temp!=null) {
for (int i = 0; i < k; i++) {
stack.push(temp.val);
temp = temp.next;
if (temp == null&&i<k-1) {
index.next = head;
return result.next;
}
}
while(!stack.isEmpty()){
index.next=new ListNode(stack.pop());
index=index.next;
head=head.next;
}
}
return result.next;
}
Java for LeetCode 025 Reverse Nodes in k-Group
原文：http://www.cnblogs.com/tonyluis/p/4474441.html