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
public class Solution {
public ListNode reverseKGroup(ListNode head, int k) {
int length = 0;
ListNode currentNode = head;
while(currentNode != null){
++length;
currentNode = currentNode.next;
}
if(k <= 1 || head == null || head.next == null || length < k)
return head;
ListNode newHead = new ListNode(0);
newHead.next = head;
currentNode = newHead;
ListNode start = head;
while(length >= k) {
ListNode end = moveNodes(start, k), nextStart = end.next;
ListNode follow = nextStart, current = start, temp = start.next;
while(current != nextStart){
current.next = follow;
follow = current;
current = temp;
if(temp != null)
temp = temp.next;
}
currentNode.next = end;
currentNode = start;
start = nextStart;
length -= k;
}
return newHead.next;
}
private ListNode moveNodes(ListNode start, int k){
ListNode end = start;
int i = 1;
while(end.next != null && i < k){
end = end.next;
++i;
}
return end;
}
}
leetcode--Reverse Nodes in k-Group,布布扣,bubuko.com
leetcode--Reverse Nodes in k-Group
原文：http://www.cnblogs.com/averillzheng/p/3807290.html