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LCR024-ReverseLinkedList.go
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135 lines (115 loc) · 2.9 KB
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package main
// LCR 024. 反转链表
// 给定单链表的头节点 head ,请反转链表,并返回反转后的链表的头节点。
// 示例 1:
// <img src="https://assets.leetcode.com/uploads/2021/02/19/rev1ex1.jpg" />
// 输入:head = [1,2,3,4,5]
// 输出:[5,4,3,2,1]
// 示例 2:
// <img src="https://assets.leetcode.com/uploads/2021/02/19/rev1ex2.jpg" />
// 输入:head = [1,2]
// 输出:[2,1]
// 示例 3:
// 输入:head = []
// 输出:[]
// 提示:
// 链表中节点的数目范围是 [0, 5000]
// -5000 <= Node.val <= 5000
// 进阶:链表可以选用迭代或递归方式完成反转。你能否用两种方法解决这道题?
import "fmt"
type ListNode struct {
Val int
Next *ListNode
}
// 打印链表
func printListNode(l *ListNode) {
if nil == l {
return
}
for {
if nil == l.Next {
fmt.Print(l.Val)
break
} else {
fmt.Print(l.Val, " -> ")
}
l = l.Next
}
fmt.Println()
}
// 数组创建链表
func makeListNode(arr []int) *ListNode {
if (len(arr) == 0) {
return nil
}
var l = (len(arr) - 1)
var head = &ListNode{arr[l], nil}
for i := l - 1; i >= 0; i-- {
var n = &ListNode{arr[i], head}
head = n
}
return head
}
/**
* Definition for singly-linked list.
* type ListNode struct {
* Val int
* Next *ListNode
* }
*/
func reverseList(head *ListNode) *ListNode {
var res *ListNode
for head != nil {
next := head.Next // 备份head.Next
head.Next = res // 更新 head.Next
res = head // 移动 new_head
head = next
//fmt.Println("new_head: ",new_head,"head: ",head)
}
return res
}
// best
func reverseList1(head *ListNode) *ListNode {
var prev *ListNode
cur := head
var next *ListNode
for cur != nil {
next = cur.Next
cur.Next = prev
prev = cur
cur = next
}
return prev
}
func main() {
l1 := makeListNode([]int{1,2,3,4,5})
fmt.Println("l1: ")
printListNode(l1)
fmt.Println("reverseList(l1): ")
printListNode(reverseList(l1))
l2 := makeListNode([]int{1,2})
fmt.Println("l2: ")
printListNode(l2)
fmt.Println("reverseList(l2): ")
printListNode(reverseList(l2))
l3 := makeListNode([]int{})
fmt.Println("l3: ")
printListNode(l3)
fmt.Println("reverseList(l3): ")
printListNode(reverseList(l3))
l1 = makeListNode([]int{1,2,3,4,5})
fmt.Println("l1: ")
printListNode(l1)
fmt.Println("reverseList1(l1): ")
printListNode(reverseList1(l1))
l2 = makeListNode([]int{1,2})
fmt.Println("l2: ")
printListNode(l2)
fmt.Println("reverseList1(l2): ")
printListNode(reverseList1(l2))
l3 = makeListNode([]int{})
fmt.Println("l3: ")
printListNode(l3)
fmt.Println("reverseList1(l3): ")
printListNode(reverseList1(l3))
}