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();

Solution

Time complexity : O(n)
Space complexity : O(1)

/**
* Definition for singly-linked list.
* struct ListNode {
*     int val;
*     ListNode *next;
*     ListNode() : val(0), next(nullptr) {}
*     ListNode(int x) : val(x), next(nullptr) {}
*     ListNode(int x, ListNode *next) : val(x), next(next) {}
* };
*/
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) {
        h = head;
    }

    /** Returns a random node's value. */
    int getRandom() {
        int ans = INT_MIN;
        int acc = 0; // accumulate
        for (ListNode* cur = h; cur != nullptr; cur=cur->next) {
            float p = (float)rand() / (float)RAND_MAX;
            if (p < (1.0/(float)++acc))
                ans = cur->val;
        }
        return ans;
    }
private:
    ListNode* h;
};

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

在一個大小未知的群體中平均取樣，在此用的是 Reservoir Sampling 演算法，其做法為：

  1. 遍歷群體中每個元素。
  2. 隨機取一機率值，若其機率落於(1/第 i 個)*100%內，則用第 i 個取代目前手上的。
  3. 全部遍歷完後，最後的還在手上的，即為所選。

其原理為：

若只看最後取到第一個的機率，
1 hit * 2 miss * 3 miss * 4 miss * ... * n miss
等於
1 * (1 - 1/i) * (1 - 1/i+1) * (1 - 1/i+2) * ... * (1 - 1/n)
等於
1 * (i-1/i) * (i/i+1) * (i+1/i+2) * ... * (n-1/n)
上下約掉後最後等於
1/n

故遍歷完整個群體，各個元素被取到的機率皆為 1/n。