# 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){
for(int i = 1; i<v.size(); i++){
while(ptr->next != NULL){
ptr = ptr->next;
}
ptr->next = new ListNode(v[i]);
}
}
class Solution {
public:
ListNode* x;
srand(time(NULL));
}
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};
cout << (ob.getRandom());
}

## Input

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

## Output

4
9
1

Updated on: 02-May-2020

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