【Leetcode】Linked List Random Node

题目链接：https://leetcode.com/problems/linked-list-random-node/

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：

链表中随即选取一个数，直接的想法：先计算链表长度len，获取[1,len]范围内的随机数作为随机到节点的下标，遍历链表获得节点值。

考虑follow up，不确定链表长度的情况下，上述做法第一步就做不了，看了tag 是 Reservoir Sampling，搜索了一下，原来这种算法解决的就是：

在不确定范围的情况下获取随机数。    参考该算法http://blog.csdn.net/yeqiuzs/article/details/52169369很容易ac了。

算法：

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

    /** Returns a random node's value. */
    public int getRandom() {
        int choice = -1;
        int idx = 0;
        ListNode p = head;
        java.util.Random r = new java.util.Random();
        while(p!=null){
            idx++;
            int rn = r.nextInt(idx);
            if(rn==idx-1){
                choice = p.val;
            }
            p = p.next;

        }
        return choice;
    }