﻿ C++ - Delete the last node of a Singly Linked List

# C++ Linked List Exercises: Delete the last node of a Singly Linked List

## C++ Linked List: Exercise-12 with Solution

Write a C++ program to delete the last node of a Singly Linked List.

Test Data:
Original list:
7 5 3 1
Remove the last node of the said list:
Updated list:
7 5 3
Again remove the last node of the said list:
Updated list:
7 5

Sample Solution:

C++ Code:

``````#include <iostream>

using namespace std;

struct  Node
{
int num;
Node *next;
}; //node constructed

int size = 0;
Node* new_Node = new Node();
new_Node->num = num;
size++;
}

{
return NULL;

return NULL;
}

// Locate the second last node
while (second_last_node->next->next != NULL)
second_last_node = second_last_node->next;

// Remove the last node from the list
delete (second_last_node->next);

// Change the next to the second last
second_last_node->next = NULL;

}

//Display all nodes
void display_all_nodes(Node* node)
{
while(node!=NULL){
cout << node->num << " ";
node = node->next;
}
}

int main()
{
cout << "Original list:\n";
cout << "\n\nRemove the last node of the said list:";
cout << "\nUpdated list:\n";
cout << "\n\nAgain remove the last node of the said list:";
cout << "\nUpdated list:\n";
cout<<endl;
return 0;
}
``````

Sample Output:

```Original list:
7 5 3 1

Remove the last node of the said list:
Updated list:
7 5 3

Again remove the last node of the said list:
Updated list:
7 5
```

Flowchart:

CPP Code Editor:

Previous C++ Exercise: Delete a middle node from a Singly Linked List.
Next C++ Exercise: Delete the nth node of a Singly Linked List from end.

What is the difficulty level of this exercise?

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## C++ Programming: Tips of the Day

How to replace all occurrences of a character in string?

std::string doesn't contain such function but you could use stand-alone replace function from algorithm header.

```#include <algorithm>
#include <string>

void some_func() {
std::string s = "example string";
std::replace( s.begin(), s.end(), 'x', 'y'); // replace all 'x' to 'y'
}
```

Ref: https://bit.ly/3niU71x