[LeetCode] 382. 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();

 

解法： 

　　因爲沒法肯定鏈表的長度，或者鏈表的長度很長，所以須要採用水塘抽樣算法。因爲限定了head必定存在，因此咱們先讓返回值res等於head的節點值，而後讓curr指向head的下一個節點，定義一個變量count，初始化爲1，若curr不爲空咱們開始循環，咱們在[0, count)中取一個隨機數，若是取出來0，那麼咱們更新res爲當前的curr的節點值，而後此時count自增一，curr指向其下一個位置，這裏其實至關於咱們維護了一個大小爲1的水塘，而後咱們隨機數生成爲0的話，咱們交換水塘中的值和當前遍歷到底值，這樣能夠保證每一個數字的機率相等，參見代碼以下：

/**
 * 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();
 */