【Leetcode】Linked List Random Node
2016-08-10 11:09
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题目链接:https://leetcode.com/problems/linked-list-random-node/
题目:
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:
思路:
链表中随即选取一个数,直接的想法:先计算链表长度len,获取[1,len]范围内的随机数作为随机到节点的下标,遍历链表获得节点值。
考虑follow up,不确定链表长度的情况下,上述做法第一步就做不了,看了tag 是 Reservoir Sampling,搜索了一下,原来这种算法解决的就是:
在不确定范围的情况下获取随机数。 参考该算法http://blog.csdn.net/yeqiuzs/article/details/52169369很容易ac了。
算法:
题目:
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();
思路:
链表中随即选取一个数,直接的想法:先计算链表长度len,获取[1,len]范围内的随机数作为随机到节点的下标,遍历链表获得节点值。
考虑follow up,不确定链表长度的情况下,上述做法第一步就做不了,看了tag 是 Reservoir Sampling,搜索了一下,原来这种算法解决的就是:
在不确定范围的情况下获取随机数。 参考该算法http://blog.csdn.net/yeqiuzs/article/details/52169369很容易ac了。
算法:
ListNode head = null; /** @param head The linked list's head. Note that the head is guanranteed 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 choice = -1; int idx = 0; ListNode p = head; java.util.Random r = new java.util.Random(); while(p!=null){ idx++; int rn = r.nextInt(idx); if(rn==idx-1){ choice = p.val; } p = p.next; } return choice; }
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