标签:Node head ListNode cur list Random List length var
package LeetCode_382 import java.util.* /** * 382. Linked List Random Node *https://leetcode.com/problems/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(); * */ class ListNode(var `val`: Int) { var next: ListNode? = null } /* solution 1: calculate the length of ListNode, then return node's value by a random num from length; * solution Follow up: Reservoir Sampling, 1/i * (1-1/(i+1)) * (1-1/(i+2)) * (1-1/(i+3)) .. (1-1/n); * */ class Solution(head: ListNode?) { var length = 0 var cur = head var head2 = head init { getLength(cur) } private fun getLength(head_: ListNode?) { var head = head_ while (head != null) { length++ head = head.next } } /** @param head The linked list's head. Note that the head is guaranteed to be not null, so it contains at least one node. */ /** Returns a random node's value. */ fun getRandom(): Int { var cur = head2 var randomNum = Random().nextInt(length) while (randomNum > 0) { randomNum-- cur = cur?.next } return cur!!.`val` } } /** * Your Solution object will be instantiated and called as such: * var obj = Solution(head) * var param_1 = obj.getRandom() */
标签:Node,head,ListNode,cur,list,Random,List,length,var 来源: https://www.cnblogs.com/johnnyzhao/p/13873333.html
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