[leetcode] 382. Linked List Random Node 解题报告

题目链接: 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();

思路: 一般的来说, 可以先计算出长度, 然后随机一个在长度范围内的值, 走到那里将值返回即可. 但是如果长度无限大, 就无法计算长度了, 这种情况下成为一个水池抽样的算法, 其原理为一个个的对元素取样, 在遍历到每个元素的时候可以有个概率选取, 或者不选取. 因为是随机选取一个数, 所以相当于水池的容量是1. 相对简单一些.

那么如何确保对于每个元素都有相等的概率呢? 这里用到了概率论的知识, 在遍历到第i个数时设置选取这个数的概率为1/i, 然后来证明一下每个数被选到的概率: 对于第一个数其被选择的概率为1/1*(1-1/2)*(1-1/3)*(1-1/4)*...*(1-1/n) = 1/n, 其中(1-1/n)的意思是不选择n的概率, 也就是选择1的概率乘以不选择其他数的概率. 对于任意一个数i来说, 其被选择的概率为1/i*(1-1/(i+1))*...*(1-1/n), 所以在每一个数的时候我们只要按照随机一个是否是i的倍数即可决定是否取当前数即可.

代码如下:

/**
* 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 guanranteed to be not null, so it contains at least one node. */
Solution(ListNode* head) {
p = head;
}

/** Returns a random node's value. */
int getRandom() {
ListNode *tem = p;
int val = p->val;
for(int i =1; tem; i++)
{
if(rand()%i==0) val = tem->val;
tem = tem->next;
}
return val;
}
private:
ListNode *p;
};

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

参考: http://www.faceye.net/search/99720.html