Leetcode 382 链表随机节点 蓄水池抽样与随机数

　　概率问题，蓄水池抽样与生成随机数都可以解决。蓄水池抽样适合数据量巨大且不知道数组集合大小时的随机抽样，因为是线性扫描每一个元素进行抽样，空间复杂度为 O(1) ,时间复杂度为O(n)。

　　蓄水池抽样算法：

class Solution {
        ListNode head;
        Random random;

        public Solution(ListNode head) {
            this.head = head;
            random = new Random(0);
        }

        public final int getRandom() {
            Integer an = null;
            int point = 1;
            ListNode node = head;
            while (node != null) {
                if (an == null) {
                    an = node.val;
                } else {
                    int ran = random.nextInt(point);
                    if (ran < 1) {
                        an = node.val;
                    }
                }
                node = node.next;
                point++;
            }
            return an == null ? 0 : an;
        }
    }

 

　　随机数抽样，需要预先知道集合大小：

class Solution {
        private ListNode head = null;
        private int size = 0;
        Random random = new Random(0);

        public Solution(ListNode head) {
            if (head != null) {
                this.head = head;
                ListNode node = head;
                while (node != null) {
                    node = node.next;
                    this.size++;
                }
            }
        }

        /**
         * Returns a random node's value.
         */
        public final int getRandom() {
            if (size <= 1) {
                return head.val;
            }
            int currentRandom = random.nextInt(size);
            return getNode(currentRandom).val;
        }

        private final ListNode getNode(int point) {
            if (point >= this.size) {
                return null;
            }
            ListNode an = head;
            while (point > 0) {
                an = an.next;
                point--;
            }
            return an;
        }

    }

　　JS 蓄水池算法：

var Solution = function (head) {
    this.head = head;
};

/**
 * Returns a random node's value.
 * @return {number}
 */
Solution.prototype.getRandom = function () {
    let point = 1;
    let an = -1;
    let node = this.head;
    while (node != null) {
        if (an == -1) {
            an = node.val;
        } else {
            let ran = parseInt(Math.random() * (point));
            if (ran < 1) {
                an = node.val;
            }
        }
        node = node.next;
        point++;
    }
    return an;
};