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();
分析：随机 采样：从 刚开始开始，首先定义一个变量，用于记录到目前为止已有多少个元素，然后利用随机算法从[0，n）中随机生成一个整数，若其等于0，则表示取当前节点，取当前节点的概率为1/n。其它的几点为1/（n-1）*（1-1/n），因此其是等概率的。
/** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { val = x; } * } */ public class Solution { ListNode head; Random randomGenerator = null; /** @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; this.randomGenerator=new Random(); } /** Returns a random node's value. */ public int getRandom() { if(head.next==null) return head.val; int ans=head.val; ListNode current=head; for(int n=1;current!=null;n++){ if(randomGenerator.nextInt(n)==0) ans=current.val; current=current.next; } return ans; } } /** * Your Solution object will be instantiated and called as such: * Solution obj = new Solution(head); * int param_1 = obj.getRandom(); */
