Go语言数据结构之单链表的实例详解
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任意类型的数据域
之前的链表定义数据域都是整型int，如果需要不同类型的数据就要用到 interface{}。
空接口 interface{}
对于描述起不到任何的作用(因为它不包含任何的method），但interface{}在需要存储任意类型的数值的时候相当有用，因为它可以存储任意类型的数值。
一个函数把interface{}作为参数，那么它可以接受任意类型的值作为参数；如果一个函数返回interface{}，那么也就可以返回任意类型的值，类似于C语言的void*类型。
package main
import "fmt"
type Node struct {
data interface{}
next *Node
}
type List struct {
head *Node
}
func (list *List) push(value interface{}) {
node := &Node{data: value}
p := list.head
if p != nil {
for p.next != nil {
p = p.next
}
p.next = node
} else {
list.head = node
}
}
func (list *List) travel() {
p := list.head
for p != nil {
fmt.Print(p.data)
p = p.next
if p != nil {
fmt.Print("->")
}
}
fmt.Println("<nil>")
}
func main() {
node := new(List)
node.push("abc")
node.push(3.142)
node.push('a')
node.push(3 + 4i)
node.push([]int{1, 2, 3})
node.push([8]byte{'a', 3: 'd'})
node.push(Node{1, &Node{2, nil}}.data)
node.push(Node{1, &Node{2, nil}}.next)
node.travel()
}
实例01
把字串中汉字除外的所有字符逐个存入链表，且数字以整型保存。
package main
import "fmt"
type Node struct {
data interface{}
next *Node
}
type List struct {
head *Node
}
func (list *List) push(value interface{}) {
node := &Node{data: value}
p := list.head
if p != nil {
for p.next != nil {
p = p.next
}
p.next = node
} else {
list.head = node
}
}
func (list *List) travel() {
p := list.head
for p != nil {
fmt.Print(p.data)
p = p.next
if p != nil {
fmt.Print("->")
}
}
fmt.Println("<nil>")
}
func main() {
node := new(List)
str := "Golang数据结构123:单链表0789"
for _, s := range str {
if s >= 48 && s < 58 {
node.push(s - 48)
} else if s < 128 {
node.push(string(s))
}
}
node.travel()
}
快慢指针
给单链表设置2个指针，其中一个指针先移动n个节点，然后同时移动这2个指针，那么当先移动的指针到达尾部时，后移动的那个指针就是倒数第 n 个节点。先移动的指针称“快指针”，后出发的指针称“慢指针”，其实一样“快”只是出发有先后。
实例02
删除链表中倒数第 n 个结点
package main
import "fmt"
type Node struct {
data interface{}
next *Node
}
type List struct {
head *Node
}
func (list *List) removNthBack(n int) {
if n > list.size() {
panic("range error: n <= List's size")
}
var fast, slow *Node
head := list.head
fast = head
slow = head
for i := 0; i < n; i++ {
fast = fast.next
}
if fast == nil {
list.head = head.next
return
}
for fast.next != nil {
fast = fast.next
slow = slow.next
}
slow.next = slow.next.next
}
func removNthBack(list *List, n int) *List {
if n > list.size() {
panic("range error: n <= List's size")
}
var fast, slow *Node
head := list.head
fast = head
slow = head
for i := 0; i < n; i++ {
fast = fast.next
}
if fast == nil {
list.head = head.next
return list
}
for fast.next != nil {
fast = fast.next
slow = slow.next
}
slow.next = slow.next.next
return list
}
func (list *List) push(value interface{}) {
node := &Node{data: value}
p := list.head
if p != nil {
for p.next != nil {
p = p.next
}
p.next = node
} else {
list.head = node
}
}
func (list *List) size() int {
length := 0
for p := list.head; p != nil; p = p.next {
length++
}
return length
}
func (list *List) travel() {
p := list.head
for p != nil {
fmt.Print(p.data)
p = p.next
if p != nil {
fmt.Print("->")
}
}
fmt.Println("<nil>")
}
func main() {
lst := new(List)
str := "12309"
for _, s := range str {
lst.push(s - 48)
}
lst.travel()
lst.removNthBack(3)
lst.travel()
lst = removNthBack(lst, 3)
lst.travel()
lst.removNthBack(2)
lst.travel()
//lst.removNthBack(10) //panic error
lst.removNthBack(2)
lst.travel()
lst.removNthBack(1)
lst.travel()
//lst.removNthBack(1) //panic error
}
反转链表
遍历一个链表，每个结点用头插法相接的新链表就是原链表的反转结果。
实例03
反转整个链表
package main
import "fmt"
type Node struct {
data interface{}
next *Node
}
type List struct {
head *Node
}
func (list *List) reverse() {
res := &List{}
for p := list.head; p != nil; p = p.next {
node := &Node{p.data, nil}
node.next = res.head
res.head = node
}
list.head = res.head
}
func (list *List) pushHead(value interface{}) {
node := &Node{data: value}
node.next = list.head
list.head = node
}
func (list *List) build(lst []interface{}) {
for i := len(lst) - 1; i >= 0; i-- {
node := &Node{data: lst[i]}
node.next = list.head
list.head = node
}
}
func (list *List) clear() {
list.head = nil
}
func (list *List) travel() {
p := list.head
for p != nil {
fmt.Print(p.data)
p = p.next
if p != nil {
fmt.Print("->")
}
}
fmt.Println("<nil>")
}
func main() {
lst := new(List)
for n := 5; n > 0; n-- {
lst.pushHead(n)
}
lst.travel()
lst.reverse()
lst.travel()
lst.clear()
lst.build([]interface{}{6.13, "/", 100000, "Hann", 1.0e-5})
lst.travel()
lst.reverse()
lst.travel()
}
实例04
反转链表的其中一段，反转从第m个结点到n个结点(其中0<m<=n<=length of List)
Input: 1->2->3->4->5->nil, m = 2, n = 4
Output: 1->4->3->2->5->nil
package main
import "fmt"
type Node struct {
data interface{}
next *Node
}
type List struct {
head *Node
}
func reverseBetween(list *List, m int, n int) *List {
list = list.Copy()
head := list.head
if head == nil || m >= n {
return list
}
if m < 1 { //防止范围左端超限
m = 1
}
node := &Node{0, head}
p := node
for i := 0; p.next != nil && i < m-1; i++ {
p = p.next
}
if p.next == nil {
return list
}
cur := p.next
for i := 0; i < n-m && cur.next != nil; i++ {
//由cur.next != nil防止范围右端超限
tmp := p.next
p.next = cur.next
cur.next = cur.next.next
p.next.next = tmp
}
list.head = node.next
return list
}
func (list *List) reverseBetween(m int, n int) {
head := list.head
if head == nil || m >= n {
return
}
if m < 1 { //防止范围左端超限
m = 1
}
node := &Node{0, head}
p := node
for i := 0; p.next != nil && i < m-1; i++ {
p = p.next
}
if p.next == nil {
return
}
cur := p.next
for i := 0; i < n-m && cur.next != nil; i++ {
//由cur.next != nil防止范围右端超限
tmp := p.next
p.next = cur.next
cur.next = cur.next.next
p.next.next = tmp
}
list.head = node.next
}
func (list *List) pushHead(value interface{}) {
node := &Node{data: value}
node.next = list.head
list.head = node
}
func (list *List) build(lst []interface{}) {
for i := len(lst) - 1; i >= 0; i-- {
node := &Node{data: lst[i]}
node.next = list.head
list.head = node
}
}
func (list *List) Copy() *List {
p := list.head
res := &List{}
if p != nil {
node := &Node{p.data, nil}
q := node
for p = p.next; p != nil; p = p.next {
q.next = &Node{p.data, nil}
q = q.next
}
res.head = node
}
return res
}
func (list *List) travel() {
p := list.head
for p != nil {
fmt.Print(p.data)
p = p.next
if p != nil {
fmt.Print("->")
}
}
fmt.Println("<nil>")
}
func main() {
list1 := new(List)
list2 := new(List)
for n := 5; n > 0; n-- {
list1.pushHead(n)
}
list1.travel()
list2 = reverseBetween(list1, 2, 4)
list2.travel()
list2 = reverseBetween(list1, 2, 3)
list2.travel()
list2 = reverseBetween(list1, 2, 5)
list2.travel()
list2 = reverseBetween(list1, 2, 6)
list2.travel()
list2 = reverseBetween(list1, 1, 6)
list2.travel()
list2 = reverseBetween(list1, 0, 3)
list2.travel()
list1.reverseBetween(1, 3)
list1.travel()
}
交换节点
实例05
链表的相邻节点两两交换位置
Given 1->2->3->4, you should return the list as 2->1->4->3.
package main
import "fmt"
type Node struct {
data interface{}
next *Node
}
type List struct {
head *Node
}
func (list *List) swapPairs() {
p := list.head
if p == nil || p.next == nil {
return
}
head := p.next
var node, node2 *Node
for p.next != nil {
cur := p.next
if node != nil && node.next != nil {
node.next = cur
}
if p.next.next != nil {
node2 = p.next.next
}
if p.next.next != nil {
p.next = node2
} else {
p.next = nil
}
cur.next = p
node = p
if p.next != nil {
p = node2
}
}
list.head = head
}
func swapPairs(list *List) *List {
list = list.Copy()
p := list.head
if p == nil || p.next == nil {
return list
}
head := p.next
var node, node2 *Node
for p.next != nil {
cur := p.next
if node != nil && node.next != nil {
node.next = cur
}
if p.next.next != nil {
node2 = p.next.next
}
if p.next.next != nil {
p.next = node2
} else {
p.next = nil
}
cur.next = p
node = p
if p.next != nil {
p = node2
}
}
list.head = head
return list
}
func (list *List) Copy() *List {
p := list.head
res := &List{}
if p != nil {
node := &Node{p.data, nil}
q := node
for p = p.next; p != nil; p = p.next {
q.next = &Node{p.data, nil}
q = q.next
}
res.head = node
}
return res
}
func (list *List) build(lst []interface{}) {
for i := len(lst) - 1; i >= 0; i-- {
node := &Node{data: lst[i]}
node.next = list.head
list.head = node
}
}
func (list *List) travel() {
p := list.head
for p != nil {
fmt.Print(p.data)
p = p.next
if p != nil {
fmt.Print("->")
}
}
fmt.Println("<nil>")
}
func main() {
list1 := new(List)
list2 := new(List)
list1.build([]interface{}{1, 2, 3, 4, 5, 6})
list1.travel()
list2 = swapPairs(list1)
list2.travel()
list2 = &List{&Node{0, nil}}
list2.head.next = list1.Copy().head
list2.travel()
list2.swapPairs()
list2.travel()
}
以上就是Go语言数据结构之单链表的实例详解的详细内容，更多关于Go语言单链表的资料请关注编程网其它相关文章！
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