[Leetcode] 382. Linked list random node linked list

Given a singly linked list, return a random node ' s value from the linked list. Each node must has the same probability of being chosen.

Follow up:
What if the linked list was extremely large and its length is unknown? Could solve this efficiently without using extra space?

Example:

Init a singly linked list [I/b]. 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 has equal probability of returning.solution.getRandom ();

Gives a list, randomly returns a node.

Solution 1: First count the length of the list, then randomly generate a position based on the length, and then traverse from the beginning to this position. But if n is big, it's not going to work.

Solution 2: Reservoir sampling reservoir sampling is a series of random algorithms that are designed to select K samples from a set of n items, where n is a large or unknown quantity, especially for cases where all n items cannot be stored in memory.

Java:

public class Solution {        ListNode head;    Random random;        Public solution (ListNode h) {        head = h;               Random = new Random ();            }        public int getrandom () {                ListNode c = head;        int r = c.val;        for (int i=1;c.next! = null;i++) {                        c = c.next;            if (Random.nextint (i + 1) = = i) R = c.val;                                }                return r;    }}

Python:

From random Import Randintclass solution (object):    def __init__ (self, head): "" "        @param head the linked list ' S head. Note that the head was guanranteed to was not null and so it contains at least one node.        : Type Head:listnode "" "        S Elf.__head = Head    # Proof of Reservoir sampling:    # https://discuss.leetcode.com/topic/53753/ Brief-explanation-for-reservoir-sampling    def getrandom (self): "" "        Returns a random node ' s value.        : Rtype:int "" "        Reservoir =-1        curr, n = self.__head, 0 while        curr:            reservoir = Curr.val if Randi NT (1, n+1) = = 1 Else reservoir            curr, n = curr.next, n+1        return Reservoir

C + +: 1

Class Solution {public:    /** @param head the linked list ' s head. Note that the head was guanranteed to was not null and so it contains at least one node. *    /solution (listnode* head) {        len = 0;        ListNode *cur = head;        This->head = head;        while (cur) {            ++len;            cur = cur->next;        }    }        /** Returns A random node ' s value. */    int getrandom () {        int t = rand ()% Len;        ListNode *cur = head;        while (t) {            --t;            cur = cur->next;        }        Return cur->val;    } Private:    int len;    ListNode *head;};

C + +: 2

Class Solution {public:    /** @param head the linked list ' s head. Note that the head was guanranteed to was not null and so it contains at least one node. *    /solution (listnode* head) {        this->head = head;    }        /** Returns A random node ' s value. *    /int getrandom () {        int res = head->val, i = 2;        ListNode *cur = head->next;        while (cur) {            int j = rand ()% i;            if (j = = 0) res = cur->val;            ++i;            cur = cur->next;        }        return res;    } Private:    listnode *head;};

　　

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398. Random Pick Index

[Leetcode] 382. Linked list random node linked list