Linked List Random Node in C++

Suppose we have a singly linked list, we have to find a random node's value from the linked list. Here each node must have the same probability of being chosen. So for example, if the list is [1,2,3], then it can return random node in range 1, 2, and 3.

To solve this, we will follow these steps −

• In the getRandom() method, do the following −

• ret := -1, len := 1, v := x

• while v is not null

  • if rand() is divisible by len, then ret := val of v

  • increase len by 1

  • v := next of v

• return ret

Example(C++)

Let us see the following implementation to get a better understanding −

 Live Demo

#include <bits/stdc++.h>
using namespace std;
class ListNode{
   public:
   int val;
   ListNode *next;
   ListNode(int data){
      val = data;
      next = NULL;
   }
};
ListNode *make_list(vector<int> v){
   ListNode *head = new ListNode(v[0]);
   for(int i = 1; i<v.size(); i++){
      ListNode *ptr = head;
      while(ptr->next != NULL){
         ptr = ptr->next;
      }
      ptr->next = new ListNode(v[i]);
   }
   return head;
}
class Solution {
   public:
   ListNode* x;
   Solution(ListNode* head) {
      srand(time(NULL));
      x = head;
   }
   int getRandom() {
      int ret = -1;
      int len = 1;
      ListNode* v = x;
      while(v){
         if(rand() % len == 0){
            ret = v->val;
         }
         len++;
         v = v->next;
      }
      return ret;
   }
};
main(){
   vector<int> v = {1,7,4,9,2,5};
   ListNode *head = make_list(v);
   Solution ob(head);
   cout << (ob.getRandom());
}

Input

Initialize list with [1,7,4,9,2,5]
Call getRandom() to get random nodes

Output

4
9
1