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();
解法：
由于无法确定链表的长度，或者链表的长度很长，因此需要采用水塘抽样算法。由于限定了head一定存在，所以我们先让返回值res等于head的节点值，然后让curr指向head的下一个节点，定义一个变量count，初始化为1，若curr不为空我们开始循环，我们在[0, count)中取一个随机数，如果取出来0，那么我们更新res为当前的curr的节点值，然后此时count自增一，curr指向其下一个位置，这里其实相当于我们维护了一个大小为1的水塘，然后我们随机数生成为0的话，我们交换水塘中的值和当前遍历到底值，这样可以保证每个数字的概率相等，参见代码如下：
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
public class Solution { private ListNode 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. */
public Solution(ListNode head) {
this.head = head;
} /** Returns a random node's value. */
public int getRandom() {
int count = 1;
int res = head.val;
ListNode curr = head.next;
while (curr != null) {
if (new Random().nextInt(++count) == 0) {
res = curr.val;
}
curr = curr.next;
}
return res;
}
} /**
* Your Solution object will be instantiated and called as such:
* Solution obj = new Solution(head);
* int param_1 = obj.getRandom();
*/