链接: K 个一组翻转链表
给你链表的头节点 head
,每 k
个节点一组进行翻转,请你返回修改后的链表。
k
是一个正整数,它的值小于或等于链表的长度。如果节点总数不是 k
的整数倍,那么请将最后剩余的节点保持原有顺序。
你不能只是单纯的改变节点内部的值,而是需要实际进行节点交换。
示例 1:
输入:head = [1,2,3,4,5], k = 2
输出:[2,1,4,3,5]
示例 2:
输入:head = [1,2,3,4,5], k = 3
输出:[3,2,1,4,5]
提示:
n
1 <= k <= n <= 5000
0 <= Node.val <= 1000
进阶: 你可以设计一个只用 O(1)
额外内存空间的算法解决此问题吗?
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode() {}
* ListNode(int val) { this.val = val; }
* ListNode(int val, ListNode next) { this.val = val; this.next = next; }
* }
*/
class Solution {
public static ListNode reverseKGroup(ListNode head, int k) {
if (k == 1) {
return head;
}
int length = getLength(head);
int index = 0;
ListNode node = head;
ListNode beforeStart = null;
ListNode start = null;
ListNode beforeNode = null;
while (node != null) {
index++;
if (index > (length - length % k)) {
assert start != null;
start.next = node;
break;
}
// 节点交换
ListNode nextNode = node.next;
if (index % k == 1) {
beforeStart = beforeNode;
start = node;
beforeNode = node;
} else if (index % k == 0) {
node.next = beforeNode;
if (beforeStart == null) {
head = node;
} else {
beforeStart.next = node;
}
start.next = nextNode;
beforeNode = start;
} else {
node.next = beforeNode;
beforeNode = node;
}
node = nextNode;
}
return head;
}
private static int getLength(ListNode head) {
int length = 0;
while (head != null) {
length++;
head = head.next;
}
return length;
}
}
整理完毕,完结撒花~