题解 | #链表的奇偶重排#
链表的奇偶重排
https://www.nowcoder.com/practice/02bf49ea45cd486daa031614f9bd6fc3
用了fast，cur1、cur2。fast负责走路，把fast、fast.Next分别发给cu1、cur2。挺别扭的，感觉没有经典思路好。
package mainimport . "nc_tools"/** type ListNode struct{* Val int* Next *ListNode* }*//*** 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可*** @param head ListNode类* @return ListNode类*/func oddEvenList(head *ListNode) *ListNode {// write code hereif head == nil {return head}dummy1 := &ListNode{}dummy2 := &ListNode{}cur1 := dummy1cur2 := dummy2fast := headfor fast != nil && fast.Next != nil {cur1.Next = fastcur1 = cur1.Nextcur2.Next = fast.Nextcur2 = cur2.Nextfast = fast.Next.Next}cur2.Next = nilcur1.Next = nilif fast == nil {cur1.Next = dummy2.Nextreturn dummy1.Next} else if fast.Next == nil {//注意当fast == nil且没有提前return时，fast.Next不存在会触发paniccur1.Next = fastcur1 = cur1.Nextcur1.Next = dummy2.Nextreturn dummy1.Next}return nil}
#后面的秋招会越来越卷吗#
