LinkedListRandomNode.cpp

// Source : https://leetcode.com/problems/linked-list-random-node/
// Author : Hao Chen
// Date   : 2016-08-24

/*************************************************************************************** 
 *
 * Given a singly linked list, return a random node's value from the linked list. Each 
 * node must have the same probability of being chosen.
 * 
 * Follow up:
 * What if the linked list is extremely large and its length is unknown to you? Could 
 * you solve this efficiently without using extra space?
 * 
 * Example:
 * 
 * // Init a singly linked list [1,2,3].
 * ListNode head = new ListNode(1);
 * head.next = new ListNode(2);
 * head.next.next = new ListNode(3);
 * Solution solution = new Solution(head);
 * 
 * // getRandom() should return either 1, 2, or 3 randomly. Each element should have 
 * equal probability of returning.
 * solution.getRandom();
 ***************************************************************************************/

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    /** @param head The linked list's head.
        Note that the head is guaranteed to be not null, so it contains at least one node. */
    Solution(ListNode* head) {
        this->head = head;
        this->len = 0;
        for(ListNode*p = head; p!=NULL; p=p->next, len++);
        srand(time(NULL));
    }
    
    /** Returns a random node's value. */
    int getRandom() {
        int pos = rand() % len;
        ListNode *p = head;
        for (; pos > 0; pos--, p=p->next);
        return p->val;
    }
    ListNode* head;
    int len;
};

/**
 * Your Solution object will be instantiated and called as such:
 * Solution obj = new Solution(head);
 * int param_1 = obj.getRandom();
 */