英文:
How to remove element from list while iterating the same list in golang
问题
我是你的中文翻译助手,以下是翻译好的内容:
我对Go语言还不熟悉。我想在遍历列表时根据条件从列表中删除元素。例如,我想从列表中删除重复的元素。以下是代码示例:
package main
import (
"container/list"
"fmt"
)
var sMap map[int]bool
func main() {
l := list.New()
l.PushFront(4)
l.PushFront(5)
l.PushFront(7)
l.PushFront(6)
l.PushFront(5)
l.PushFront(4)
l.PushFront(5)
l.PushFront(7)
l.PushBack(9)
l = removeDuplicate(l)
for e := l.Front(); e != nil; e = e.Next() {
fmt.Println(e.Value)
}
}
func removeDuplicate(l *list.List) *list.List {
sMap = make(map[int]bool)
for e := l.Front(); e != nil; e = e.Next() {
m := e.Value.(int)
fmt.Println("VALUE : ", m)
if sMap[m] == true {
fmt.Println("Deleting ", e.Value)
l.Remove(e)
} else {
fmt.Println("Adding New Entry", e.Value)
sMap[m] = true
}
}
return l
}
上述代码只在第一次删除后遍历了列表。我试图在遍历同一个列表时删除元素,这就是它不起作用的原因。有人能否建议一种在Go语言中进行列表迭代的方法?
英文:
I am new to go language. I would like to remove elements from the list while iterating the list based on a condition in go language. For example I want remove the duplicate elements from the list. Code is given below.
package main
import (
"container/list"
"fmt"
)
var sMap map[int]bool
func main() {
l := list.New()
l.PushFront(4)
l.PushFront(5)
l.PushFront(7)
l.PushFront(6)
l.PushFront(5)
l.PushFront(4)
l.PushFront(5)
l.PushFront(7)
l.PushBack(9)
l = removeDuplicate(l)
for e := l.Front(); e != nil; e = e.Next() {
fmt.Println(e.Value)
}
}
func removeDuplicate(l *list.List) *list.List {
sMap = make(map[int]bool)
for e := l.Front(); e != nil; e = e.Next() {
m := e.Value.(int)
fmt.Println("VALUE : ", m)
if sMap[m] == true {
fmt.Println("Deleting ", e.Value)
l.Remove(e)
} else {
fmt.Println("Adding New Entry", e.Value)
sMap[m] = true
}
}
return l
}
The above code iterates through the list only till the first removal. I am trying to remove the element while iterating through the same list. That is the reason why it is not working. Could anyone suggest an list iterator in golang?
答案1
得分: 22
如果从列表中删除了e,那么在下一次循环中调用e.Next()将返回nil。因此,在删除e之前,需要将e.Next()赋值给next。以下是通过迭代清除所有元素的示例(在list_test.go中):
// 通过迭代清除所有元素
var next *Element
for e := l.Front(); e != nil; e = next {
next = e.Next()
l.Remove(e)
}
同样的模式可以应用于以下问题:
package main
import (
"container/list"
"fmt"
)
var sMap map[int]bool
func main() {
l := list.New()
l.PushFront(4)
l.PushFront(5)
l.PushFront(7)
l.PushFront(6)
l.PushFront(5)
l.PushFront(4)
l.PushFront(5)
l.PushFront(7)
l.PushBack(9)
l = removeDuplicate(l)
for e := l.Front(); e != nil; e = e.Next() {
fmt.Println(e.Value)
}
}
func removeDuplicate(l *list.List) *list.List {
sMap = make(map[int]bool)
var next *list.Element
for e := l.Front(); e != nil; e = next {
m := e.Value.(int)
next = e.Next()
fmt.Println("VALUE : ", m)
if sMap[m] == true {
fmt.Println("Deleting ", e.Value)
l.Remove(e)
} else {
fmt.Println("Adding New Entry", e.Value)
sMap[m] = true
}
}
return l
}
输出结果:
VALUE : 7
Adding New Entry 7
VALUE : 5
Adding New Entry 5
VALUE : 4
Adding New Entry 4
VALUE : 5
Deleting 5
VALUE : 6
Adding New Entry 6
VALUE : 7
Deleting 7
VALUE : 5
Deleting 5
VALUE : 4
Deleting 4
VALUE : 9
Adding New Entry 9
7
5
4
6
9
英文:
If e is removed from the list then call of e.Next() in the next loop will return nil. Therefore, need to assign e.Next() to the next before deleting e. Here is the example to clear all elements by iterating (in list_test.go)
// Clear all elements by iterating
var next *Element
for e := l.Front(); e != nil; e = next {
next = e.Next()
l.Remove(e)
}
Same pattern can be applied to the question as following;
package main
import (
"container/list"
"fmt"
)
var sMap map[int]bool
func main() {
l := list.New()
l.PushFront(4)
l.PushFront(5)
l.PushFront(7)
l.PushFront(6)
l.PushFront(5)
l.PushFront(4)
l.PushFront(5)
l.PushFront(7)
l.PushBack(9)
l = removeDuplicate(l)
for e := l.Front(); e != nil; e = e.Next() {
fmt.Println(e.Value)
}
}
func removeDuplicate(l *list.List) *list.List {
sMap = make(map[int]bool)
var next *list.Element
for e := l.Front(); e != nil; e = next {
m := e.Value.(int)
next = e.Next()
fmt.Println("VALUE : ", m)
if sMap[m] == true {
fmt.Println("Deleting ", e.Value)
l.Remove(e)
} else {
fmt.Println("Adding New Entry", e.Value)
sMap[m] = true
}
}
return l
}
Output
VALUE : 7
Adding New Entry 7
VALUE : 5
Adding New Entry 5
VALUE : 4
Adding New Entry 4
VALUE : 5
Deleting 5
VALUE : 6
Adding New Entry 6
VALUE : 7
Deleting 7
VALUE : 5
Deleting 5
VALUE : 4
Deleting 4
VALUE : 9
Adding New Entry 9
7
5
4
6
9
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