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();
给一个链表，随机返回一个节点。
解法1：先统计出链表的长度，然后根据长度随机生成一个位置，然后从开头遍历到这个位置。但如果n很大就不好处理了。
解法2: 水塘抽样 Reservoir sampling，是一系列的随机算法，其目的在于从包含n个项目的集合S中选取k个样本，其中n为一很大或未知的数量，尤其适用于不能把所有n个项目都存放到内存的情况。
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 randint
class Solution(object):
def __init__(self, head):
"""
@param head The linked list's head. Note that the head is guanranteed to be not null, so it contains at least one node.
:type head: ListNode
"""
self.__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 randint(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 is guanranteed to be not null, 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 is guanranteed to be not null, 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;
};
类似题目：
398. Random Pick Index