【LeetCode】382.Linked List Random Node（medium）解题报告

【LeetCode】382.Linked List Random Node（medium）解题报告

tags: Reservoir Sampling

题目地址：https://leetcode.com/problems/linked-list-random-node/description/
题目描述：

  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.
  What if the linked list is extremely large and its length is unknown to you? Could you solve this efficiently without using extra space?

Examples:

// 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();

Solutions:

/** * Definition for singly-linked list. * public class ListNode { *     int val; *     ListNode next; *     ListNode(int x) { val = x; } * } *//*lottery gamelist:    1->2->3...index:   0->1->2...cur size:[0,3)winner:3*/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 winner = head.val;        ListNode cur = head;        for(int i = 1 ; cur.next!=null; i++){            cur = cur.next;            if(getRandomNum(i+1)==i){                winner = cur.val;            }        }        return winner;    }    public int getRandomNum(int n){        return (int)(Math.random()*n);        //Math.random()*n -> [0,1)*n -> [0,n)    }}/** * Your Solution object will be instantiated and called as such: * Solution obj = new Solution(head); * int param_1 = obj.getRandom(); */

  重点就是相同的可能性。还是要记住random函数返回值范围。

Date：2017年10月22日