382. Linked List Random Node

Q:

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();

A:

就是每次往后 有 (n-1)/n 的几率 保留该值，往后flip

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    ListNode header;
    /** @param head The linked list's head.
        Note that the head is guaranteed to be not null, so it contains at least one node. */
    public Solution(ListNode head) {
        header = new ListNode(0);
        header.next = head;
    }
    
    /** Returns a random node's value. */
    public int getRandom() {
        ListNode cur = header.next;
        int count = 0, res = 0;
        while(cur !=null){
            count++;
            double rand = Math.random();
            if( rand < (1.0/count) ){
                res= cur.val;
            }
            cur = cur.next;
        }
        return res;
    }
}

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

Mistakes: