Created solution code for "Binary Lined list to integer"

LinkedList/Convert_binary_to_linked_list_int.cpp

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+/*
+Problem Statement
+Given head which is a reference node to a singly-linked list. The value of each node in the linked list is either 0 or 1. The linked list holds the binary representation of a number.
+
+Return the decimal value of the number in the linked list.
+
+Input Format
+the first line contains n the number of nodes in the given linked list
+
+the second line contains n space separated binary integers to be in the list.
+
+Output Format
+Calculate and return the number formed by using the input list
+
+Constraints
+The Linked List is not empty.
+Number of nodes will not exceed 30.
+Each node's value is either 0 or 1.
+Sample Testcase 0
+Testcase Input
+3
+1 0 1
+Testcase Output
+5
+*/
+
+
+// code
+#include <iostream>
+
+struct ListNode {
+    int val;
+    ListNode* next;
+    ListNode(int x) : val(x), next(nullptr) {}
+};
+
+int getDecimalValue(ListNode* head) {
+    int decimalValue = 0;
+    while (head != nullptr) {
+        decimalValue = (decimalValue << 1) | head->val;
+        head = head->next;
+    }
+    return decimalValue;
+}
+
+// Function to create a linked list from an array of values
+ListNode* createLinkedList(int values[], int n) {
+    if (n == 0) return nullptr;
+    ListNode* head = new ListNode(values[0]);
+    ListNode* current = head;
+    for (int i = 1; i < n; ++i) {
+        current->next = new ListNode(values[i]);
+        current = current->next;
+    }
+    return head;
+}
+
+int main() {
+    int n;
+    std::cin >> n;
+    int binaryValues[n];
+    for (int i = 0; i < n; ++i) {
+        std::cin >> binaryValues[i];
+    }
+
+    ListNode* head = createLinkedList(binaryValues, n);
+    int decimalValue = getDecimalValue(head);
+
+    std::cout << decimalValue << std::endl;
+
+    return 0;
+}