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

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();

思路：

最近的题目感觉和随机数相关的比较多，不过这个确实是面试的趋势。对于本题目，我们首先看看允许使用额外空间的解法，然后再看看如果不允许使用额外空间，怎么利用Reservior Sampling方法获得随机结点，这应该是本题目考查的重点。

1、使用额外空间：我们将单链表的结点值都拷贝到一个数组中，然后产生一个随机数索引，返回随机索引对应的值即可。这种做法的空间复杂度是O(n)；构造函数的时间复杂度是O(n)，getRandom函数的时间复杂度是O(1)。另外一个变种是我们在构造函数中计算出整个单链表的长度length，然后在getRandom方法中，随机产生一个介于[0, length]之间的数x，然后遍历单链表，返回第x个结点。这种做法的空间复杂度可以降低到O(1)，但是getRandom的时间复杂度是O(n)。以上两种做法在length变为无限大的时候都会失效。

2、不使用额外空间：先来介绍一下Reservior Sampling方法，这种采样非常适合于流数据。怎么做呢？假设目前我们已经处理了n个结点，而当前选中的采样点是x（1 <= x <= n）；这样当我们遇到第n+1个结点的时候，让x以1 / (n + 1)的概率转移到第n+1个结点。可以用概率论的知识证明当处理完所有结点后，每个结点被采样的概率是相同的。具体实现请见下面的代码。

代码：

1、使用额外空间：

/*** 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 guaranteed to be not null, so it contains at least one node. */Solution(ListNode* head) {ListNode* node = head;while(node != NULL) {vec.push_back(node->val);node = node->next;}}/** Returns a random node's value. */int getRandom() {if(vec.size() == 0) {return -1;}return vec[rand() % vec.size()];}
private:vector<int> vec;
};/*** Your Solution object will be instantiated and called as such:* Solution obj = new Solution(head);* int param_1 = obj.getRandom();*/

2、不使用额外空间：

/*** 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 guaranteed to be not null, so it contains at least one node. */Solution(ListNode* head) {random_list = head;}/** Returns a random node's value. */int getRandom() {ListNode* temp = random_list;int value = random_list->val;for (int i = 1; temp != NULL; ++i) {if (rand() % i == 0) {      // the propability is 1 / ivalue = temp->val;}temp = temp->next;}return value;}
private:ListNode* random_list;
};/*** Your Solution object will be instantiated and called as such:* Solution obj = new Solution(head);* int param_1 = obj.getRandom();*/

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