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Golang program to delete the ith index node, when index is the out of range in the linked list.
Example
Approach to solve this problem
Step 1 − Define a method that accepts the head of a linked list.
Step 2 − If head == nil, return the head.
Step 3 − When index == 0, then return head.next
Step 4 − Else, iterate the given linked list from its head.
Step 5 − If index i matches with the given index (to be deleted), then delete that node.next, break the loop.
Step 6 − Return, at the end of the loop.
Example
package main
import "fmt"
type Node struct {
value int
next *Node
}
func NewNode(value int, next *Node) *Node{
var n Node
n.value = value
n.next = next
return &n
}
func TraverseLinkedList(head *Node){
temp := head
for temp != nil {
fmt.Printf("%d ", temp.value)
temp = temp.next
}
fmt.Println()
}
func DeleteKthIndexNode(head *Node, index int) *Node{
if head == nil{
return head
}
if index == 0{
head = head.next
return head
}
i := 1
temp := head
for temp != nil{
if i == index{
temp.next = temp.next.next
}
i++
temp = temp.next
}
return head
}
func main(){
head := NewNode(30, NewNode(10, NewNode(40, NewNode(40, nil))))
fmt.Printf("Input Linked list is: ")
TraverseLinkedList(head)
index := 5
head = DeleteKthIndexNode(head, index)
fmt.Printf("After Deletion of %dth index node, Linked List is: ", index)
TraverseLinkedList(head)
}Output
Input Linked list is: 30 10 40 40 After Deletion of 5th index node, Linked List is: 30 10 40 40
Updated on: 18-Mar-2021
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