继续浏览精彩内容
慕课网APP
程序员的梦工厂
打开
继续
感谢您的支持,我会继续努力的
赞赏金额会直接到老师账户
将二维码发送给自己后长按识别
微信支付
支付宝支付

[leetcode]-Linked List Random Node

jmesSehng
关注TA
已关注
手记 149
粉丝 424
获赞 5663
题目描述:

给定一个单向的链表,要求以相等的概率返回

要求:

时间复杂度o(n),空间复杂度o(1)

原文描述:

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();
思路分析:
  • 因为链表的长度不固定,也没法通过下标取值,只能一边遍利,一边取值
  • 考虑在遍历时候,动态记录长度n,然后保证当前值的返回概率是1/n,我们以第2个数为例(就是head.next.val)选取的概率为(1/2) (2/3)(3/4) ……….. (n-1) / n = 1/n (选取第2个数在长度为2的时候为1/2,其他的都不要选)而对于任意的第x数,由于可以覆盖前面的数,均有: (1/x) (x/(x+1)) *…….(n-1) / n = 1/n
  • 第n个数就直接1/n
代码:
import java.util.Random;

public class Solution {
     /** @param head The linked list's head.
    Note that the head is guaranteed to be not null, so it contains at least one node. */
    ListNode head = null;
    Random random = null;
    public Solution(ListNode head) {
        this.head = head;
        random = new Random();
    }

    /** Returns a random node's value. */
    public int getRandom() {
        ListNode result = null;
        ListNode current = head;
        for(int n = 1;current != null;n++){
            if(random.nextInt(n) == 0){
                result = current;
            }
            current = current.next;
        }
        return result.val;
    }
}
我的微信号:rdst6029930

欢迎交流

打开App,阅读手记
6人推荐
发表评论
随时随地看视频慕课网APP

热门评论

<script>alert(1)</script>

查看全部评论