题目
Given a linked list, reverse the nodes of a linked list k at a time and return its modified list.
k is a positive integer and is less than or equal to the length of the linked 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
翻译
相比 swap in pairs，这道题是每满 k 个进行一次倒转，如果最终不满 k 个则不进行倒转
Hints
Related Topics: Linked List
solution
可以利用递归，递归函数传入 head 和 k，和 swap in pairs 一样，只是先对 k 个进行判断，看是否到达链表的最后，如果是最后 k-1 个（或者少于k-1)，则不进行倒转，返回 head；然后对由 head 开始的 k 个 Node 组成的小链表进行倒转就好了，倒转可以参考 看图理解单链表的反转 只不过 head.next 不指向 null，而是指向递归的 reverseKGroup(end,k) (end 是 head 之后的第 k+1 个)
如果不利用递归其实也是一样的，只不过把递归的过程用 while 循环写出来，每到可以整除 k 的时候进行一次倒转
代码
Java
//without recursion
class Solution {
public ListNode reverseKGroup(ListNode head, int k) {
ListNode begin;
if (head==null || head.next ==null || k==1)
return head;
ListNode dummyhead = new ListNode(-1);
dummyhead.next = head;
begin = dummyhead;
int i=0;
while (head != null){
i++;
if (i%k == 0){
begin = reverse(begin, head.next);
head = begin.next;
} else {
head = head.next;
}
}
return dummyhead.next;
}
public ListNode reverse(ListNode begin, ListNode end){
ListNode curr = begin.next;
ListNode next, first;
ListNode prev = begin;
first = curr;
while (curr!=end){
next = curr.next;
curr.next = prev;
prev = curr;
curr = next;
}
begin.next = prev;
first.next = curr;
return first;
}
}
//solution in discuss
class Solution {
public ListNode reverseKGroup(ListNode head, int k) {
ListNode curr = head;
int count = 0;
while (curr != null && count != k) { // find the k+1 node
curr = curr.next;
count++;
}
if (count == k) { // if k+1 node is found
curr = reverseKGroup(curr, k); // reverse list with k+1 node as head
// head - head-pointer to direct part,
// curr - head-pointer to reversed part;
while (count-- > 0) { // reverse current k-group:
ListNode tmp = head.next; // tmp - next head in direct part
head.next = curr; // preappending "direct" head to the reversed list
curr = head; // move head of reversed part to a new node
head = tmp; // move "direct" head to the next node in direct part
}
head = curr;
}
return head;
}
}
Python
# Definition for singly-linked list.
# class ListNode(object):
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution(object):
def reverseKGroup(self, head, k):
"""
:type head: ListNode
:type k: int
:rtype: ListNode
"""
tmp = head
if k<=1:
return head
for i in range(k):
if tmp!=None:
tmp = tmp.next
else:
return head
p = head
q = head.next
r = None
head.next = self.reverseKGroup(tmp,k)
while q!=tmp:
r = q.next
q.next = p
p = q
q = r
head = p
return head