25.以組的形式反轉節點
25. Reverse Nodes in k-Group
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.
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
Note:
- Only constant extra memory is allowed.
- You may not alter the values in the list’s nodes, only nodes itself may be changed.
尾插法
每次先掃描k個節點,如果k個節點出現了null的節點,那麼說明剩下的節點不足k個,所以不需要反轉。直接保留。如果沒有出現null節點,就對這k個節點進行反轉。這裏使用尾插法進行反轉。尾插法即先固定頭部,新添加的節點在後面添加。
以1->2->3->4
,k爲3的情況爲例。它的變換過程如下。
3->1->2->4
3->2->1->4
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
class Solution {
public ListNode reverseKGroup(ListNode head, int k) {
if(head == null || k <= 1|| head.next == null){
return head;
}
ListNode first = new ListNode(0);
first.next = head;
ListNode pre = first;
ListNode cur = pre.next;
ListNode post;
ListNode end = first;
while(true){
int i = 0;
while(i < k){
end = end.next;
if(end== null){
return first.next;
}
i++;
}
for(int j = 0; j < k -1; j++){
post = cur.next;
cur.next = post.next;
post.next = pre.next;
pre.next = post;
}
pre = cur;
end = cur;
cur = pre.next;
}
}
}
頭插法
和上面的思想差不多,而這裏使用的是頭插法。頭插法與尾插法相反,先固定尾部,然後新的節點在頭部添加。
以1->2->3->4
,k爲3的情況爲例。它的變換過程如下。
2->3->1->4
3->2->1->4
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
class Solution {
public ListNode reverseKGroup(ListNode head, int k) {
int i = 0;
ListNode p = head;
while(i < k && p!=null){
p = p.next;
i++;
}
if(i <k){
return head;
}
ListNode pre = reverseKGroup(p,k);
i = 0;
p = head;
while(i !=k){
ListNode next = p.next;
p.next = pre;
pre = p;
p = next;
i++;
}
return pre;
}
}