卖萌的弱渣

I am stupid, I am hungry.

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:

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// 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

Proof

Java

(Linked-List-Random-Node.java) download
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/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {

    /** @param head The linked list's head.
        Note that the head is guaranteed to be not null, so it contains at least one node. */
    ListNode head = null;
    Random randomGenerator = null;
    public Solution(ListNode head) {
        this.head = head;
        this.randomGenerator = new Random();
    }

    /** Returns a random node's value. */
    public int getRandom() {
        int ret=0;
        ListNode cur = this.head;
        for(int i=1;cur!=null;i++){
            if(randomGenerator.nextInt(i)==0)
                ret = cur.val;
            cur = cur.next;
        }
    return ret;
    }
}

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

Python

  • 我们以第2个数为例（就是head.next.val）

  • 在从[0,n]遍历一边的过程中被选取的概率为(1/2)* （2/3）*（3/4）* ……….. (n-1) / n = 1/n （选取第2个数在长度为2的时候为1/2，其他的都不要选)

  • 此一类推，任何一个数x在遍历一边中被选到的概率都是1/n

(Linked-List-Random-Node.py) download
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# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution(object):

    def __init__(self, head):
        """
        @param head The linked list's head.
        Note that the head is guaranteed to be not null, so it contains at least one node.
        :type head: ListNode
        """
        self.head = head

    def getRandom(self):
        """
        Returns a random node's value.
        :rtype: int
        """
        i = 0
        ans = 0
        head = self.head
        while head:
            # randint 是在[0,i]中生成随机数
            # roll到0才算数
            if random.randint(0,i) == 0:
                ans = head.val
            # head 一直在前进
            head = head.next
            i = i+
        return ans

# Your Solution object will be instantiated and called as such:
# obj = Solution(head)
# param_1 = obj.getRandom()