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Move all zeros to the front of the linked list in C++
Given a linked list with random integers and zeroes. We have to move all the zeroes to the front of the linked list. Let's see an example.
Input
3 -> 0 -> 1-> 0 -> 0 -> 1 -> 0 -> 0 -> 3 -> NULL
Output
0->0->0->0->0->3->1->1->3->NULL
Algorithm
- Initialise the linked list.
- Return if the linked list is empty or it has single node.
- Initialise two nodes with second node and first node respectively to track current and previous nodes.
- Iterate over the linked list until we reach the end.
- If the current node is 0, then make it new head.
- Update the values of current and previous nodes variables.
- Making the node new head will move it to the front.
- Update the next value of the new head with the previous head.
Implementation
Following is the implementation of the above algorithm in C++
#include <bits/stdc++.h>
using namespace std;
struct Node {
int data;
struct Node *next;
};
void addNewNode(struct Node **head, int data) {
struct Node *newNode = new Node;
newNode->data = data;
newNode->next = *head;
*head = newNode;
}
void moveZeroes(struct Node **head) {
if (*head == NULL) {
return;
}
struct Node *temp = (*head)->next, *prev = *head;
while (temp != NULL) {
if (temp->data == 0) {
Node *current = temp;
temp = temp->next;
prev->next = temp;
current->next = *head;
*head = current;
}else {
prev = temp;
temp = temp->next;
}
}
}
void printLinkedList(struct Node *head) {
while (head != NULL) {
cout << head->data << "->";
head = head->next;
}
cout << "NULL" << endl;
}
int main() {
struct Node *head = NULL;
addNewNode(&head, 3);
addNewNode(&head, 0);
addNewNode(&head, 1);
addNewNode(&head, 0);
addNewNode(&head, 0);
addNewNode(&head, 1);
addNewNode(&head, 0);
addNewNode(&head, 0);
addNewNode(&head, 3);
moveZeroes(&head);
printLinkedList(head);
return 0;
}Output
If you run the above code, then you will get the following result.
0->0->0->0->0->3->1->1->3->NULL
Updated on: 25-Oct-2021
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