382. Linked List Random Node (Medium)

https://leetcode.com/problems/linked-list-random-node/

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();

Hints

Solutions

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    ListNode head;
    Random random;

    /** @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;
        this.random = new Random();
    }

    /** Returns a random node's value. */
    public int getRandom() {
        ListNode curr = head;
        ListNode rst = null;
        for (int i = 1; curr != null; i++) {
            // random.nextInt(n) between 0 (inclusive) and n (exclusive)
            // determine whether the ith is put into the pool or not.
            // For the nth target, i is n. 
            // Then the probability that rnd.nextInt(i)==0 is 1/n. 
            // Thus, the probability that return nth target is 1/n.
            if (random.nextInt(i) == 0) {
                rst = curr;
            }
            curr = curr.next;
        }
        return rst.val;
    }
}


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