LeetCode 382. Linked List Random Node

原题链接在这里：https://leetcode.com/problems/linked-list-random-node/

题目：

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();

题解：

First assign head value to res. count = 1.

Then while current node next != null, move cur to next,  pick random number within [0, ++count).

If it is equal to 0, update res.

Time Complexity: getRandom, O(n). n is length of list.

Space: O(1).

AC Java:

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    ListNode head;
    Random rand;
    
    /** @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) {
        this.head = head;
        rand = new Random();
    }
    
    /** Returns a random node's value. */
    public int getRandom() {
        ListNode cur = head;
        int res = cur.val;
        int count = 1;
        while(cur.next != null){
            cur = cur.next;
            if(rand.nextInt(++count) == 0){
                res = cur.val;
            }
        }
        
        return res;
    }
}

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

类似Random Pick Index.