LeetCode-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: 

// 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();

Analysis:
One way is reservoir sampling: https://en.wikipedia.org/wiki/Reservoir_sampling
Solution:
 1 /**
 2  * Definition for singly-linked list.
 3  * public class ListNode {
 4  *     int val;
 5  *     ListNode next;
 6  *     ListNode(int x) { val = x; }
 7  * }
 8  */
 9 public class Solution {
10     ListNode first;
11 
12     /** @param head The linked list's head.
13         Note that the head is guaranteed to be not null, so it contains at least one node. */
14     public Solution(ListNode head) {
15         first = head;
16     }
17     
18     /** Returns a random node's value. */
19     public int getRandom() {
20         ListNode target = first;
21         int val = target.val;
22         Random engine = new Random();
23         for (int i=1;target!=null;i++){
24             if (engine.nextInt(i)==0){
25                 val = target.val;
26             }
27             target = target.next;
28         }
29         return val;
30     }
31 }
32 
33 /**
34  * Your Solution object will be instantiated and called as such:
35  * Solution obj = new Solution(head);
36  * int param_1 = obj.getRandom();
37  */

Solution 2: Regular answer

 1 /**
 2  * Definition for singly-linked list.
 3  * public class ListNode {
 4  *     int val;
 5  *     ListNode next;
 6  *     ListNode(int x) { val = x; }
 7  * }
 8  */
 9 public class Solution {
10     
11     Map<Integer,Integer> indexMap;
12     int index;
13 
14     /** @param head The linked list's head.
15         Note that the head is guaranteed to be not null, so it contains at least one node. */
16     public Solution(ListNode head) {
17         indexMap = new HashMap<Integer,Integer>();
18         index = 0;
19         ListNode cur = head;
20         while (cur!=null){
21             indexMap.put(index,cur.val);
22             index++;
23             cur = cur.next;
24         }
25     }
26     
27     /** Returns a random node's value. */
28     public int getRandom() {
29        Random engine = new Random();
30        int ind = engine.nextInt(index);
31        return indexMap.get(ind);
32     }
33 }
34 
35 /**
36  * Your Solution object will be instantiated and called as such:
37  * Solution obj = new Solution(head);
38  * int param_1 = obj.getRandom();
39  */

 




posted @ 2016-08-17 13:39  LiBlog  阅读(230)  评论(0)  编辑  收藏  举报