leetcode:K 个一组翻转链表(没弄懂)
作者:互联网
思路:
1.递归,每次处理k个,不足k个不处理
2.对k个的处理用reverse进行
3.reverse实际上由k+1个,通过prev,first和temp来逐次挪动
src
/**
* Definition for singly-linked list.
* type ListNode struct {
* Val int
* Next *ListNode
* }
*/
func reverseKGroup(head *ListNode, k int) *ListNode {
node := head
for i := 0; i < k; i++ {
if node == nil {
return head
}
node = node.Next
}
newHead := reverse(head, node)
head.Next = reverseKGroup(node, k)
return newHead
}
func reverse(first, last *ListNode) *ListNode {
prev := last
for first != last {
tmp := first.Next
first.Next = prev
prev = first
first = tmp
}
return prev
}
总结:
思路get了,细节没get
标签:node,head,ListNode,弄懂,Next,链表,prev,leetcode,first 来源: https://blog.csdn.net/weixin_40986490/article/details/120734226