Question: Reversing a singly linked list.
输入: 1->2->3->4->5->NULL 输出: 5->4->3->2->1->NULL
First of all, let’s get to know the data structure of the linked list:
There are two elements in the linked list node:
- Value
- Pointer
type ListNode struct {
Val int
Next *ListNode
}
Next points to the next node
Then this question is actually to point the pointer to the previous node
| Number of position changes | pre | cur | whole |
|---|---|---|---|
| 0 | nil | 1->2->3->4->5 | 1->2->3->4->5 |
| 1 | 1->nil | 2->-3>->4->5 | 2->3->4->5->1->nil |
| 2 | 2->1->nil | 3->4->5 | 3->4->5->2->1->nil |
| 3 | 3->2->1->nil | 4->5 | 4->5->3 ->2->1->nil |
| 4 | ##4->3->2->1->nil5 | 5->4->3->2->1->nil |
- pre is the frontmost element of cur (pre = cur)
- cur is the linked list element behind the current position (cur = cur.Next)
- cur .Next must be connected to pre (cur.Next = pre)
package main
import "fmt"
//链表节点
type ListNode struct {
Val int
Next *ListNode
}
//反转链表的实现
func reversrList(head *ListNode) *ListNode {
cur := head
var pre *ListNode = nil
for cur != nil {
pre, cur, cur.Next = cur, cur.Next, pre //这句话最重要
}
return pre
}
func main() {
head := new(ListNode)
head.Val = 1
ln2 := new(ListNode)
ln2.Val = 2
ln3 := new(ListNode)
ln3.Val = 3
ln4 := new(ListNode)
ln4.Val = 4
ln5 := new(ListNode)
ln5.Val = 5
head.Next = ln2
ln2.Next = ln3
ln3.Next = ln4
ln4.Next = ln5
pre := reversrList(head)
fmt.Println(pre)
}
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