K 个一组翻转链表(递归,Kotlin)
K 个一组翻转链表(递归,Kotlin)
25. K 个一组翻转链表
给你一个链表，每 k 个节点一组进行翻转，请你返回翻转后的链表。
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。
示例：
给你这个链表：1->2->3->4->5
当 k = 2 时，应当返回: 2->1->4->3->5
当 k = 3 时，应当返回: 3->2->1->4->5
说明：
你的算法只能使用常数的额外空间。 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。
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.
解题思路
1.分组反转链表操作 2.边界处理
代码
class Solution {
fun reverseKGroup(head: ListNode?, k: Int): ListNode? {
if (head?.next == null) {
return head
}
// 找到tail的地址
var tail = head
for (i in 0..(k - 1)) {
// 当这一组链表节点数小于k,直接返回
if (tail == null) {
return head
}
// 指针逐一后移,一直移到k-1位置=tail
tail = tail.next
}
// 反转该组
var newHead = reverse(head, tail)
// head.next = 下一组的 newHead
head.next = reverseKGroup(tail, k)
// 返回当前组的newHead
return newHead
}
fun reverse(head: ListNode?, tail: ListNode?): ListNode? {
var cur = head
// cur pre
var pre: ListNode? = null
// cur next
var next: ListNode?
// cur->cur.next
while (cur != tail) {
// next持有cur的next节点
next = cur?.next
// 反转cur节点的next为pre
cur?.next = pre
// 指针向后移到下一个节点
pre = cur
cur = next
}
// 返回newHead
return pre
}
}
class ListNode(var value: Int) {
var next: ListNode? = null
override fun toString(): String {
return "ListNode(value=$value, next=$next)"
}
}测试函数:
fun main() {
run {
val listNode1 = ListNode(1)
val listNode2 = ListNode(2)
val listNode3 = ListNode(3)
val listNode4 = ListNode(4)
val listNode5 = ListNode(5)
listNode1.next = listNode2
listNode2.next = listNode3
listNode3.next = listNode4
listNode4.next = listNode5
listNode5.next = null
val ans = reverseKGroup(listNode1, 2)
println(ans)
}
run {
val listNode1 = ListNode(1)
val listNode2 = ListNode(2)
val listNode3 = ListNode(3)
val listNode4 = ListNode(4)
val listNode5 = ListNode(5)
listNode1.next = listNode2
listNode2.next = listNode3
listNode3.next = listNode4
listNode4.next = listNode5
listNode5.next = null
val ans = reverseKGroup(listNode1, 3)
println(ans)
}
}输出:
ListNode(value=2, next=ListNode(value=1, next=ListNode(value=4, next=ListNode(value=3, next=ListNode(value=5, next=null))))) ListNode(value=3, next=ListNode(value=2, next=ListNode(value=1, next=ListNode(value=4, next=ListNode(value=5, next=null)))))
相似的题型
反转链表
Reverse a singly linked list. Example:
Input: 1->2->3->4->5->NULL Output: 5->4->3->2->1->NULL Follow up:
A linked list can be reversed either iteratively or recursively. Could you implement both? https://leetcode-cn.com/explore/interview/card/bytedance/244/linked-list-and-tree/1038/
源代码:
/**
* Example:
* var li = ListNode(5)
* var v = li.`val`
* Definition for singly-linked list.
* class ListNode(var `val`: Int) {
* var next: ListNode? = null
* }
*/
class Solution {
fun reverseList(head: ListNode?): ListNode? {
var cur = head
var pre:ListNode? = null
var next:ListNode? = null
while(cur!=null){
next = cur.next
cur.next = pre
pre = cur
cur = next
}
return pre
}
}参考资料
https://leetcode-cn.com/problems/reverse-nodes-in-k-group/ https://leetcode-cn.com/submissions/detail/67275479/