62 lines
1.8 KiB
C++
62 lines
1.8 KiB
C++
// Source : https://leetcode.com/problems/linked-list-random-node/
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// Author : Hao Chen
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// Date : 2016-08-24
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/***************************************************************************************
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*
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* Given a singly linked list, return a random node's value from the linked list. Each
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* node must have the same probability of being chosen.
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*
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* Follow up:
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* What if the linked list is extremely large and its length is unknown to you? Could
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* you solve this efficiently without using extra space?
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*
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* Example:
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*
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* // Init a singly linked list [1,2,3].
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* ListNode head = new ListNode(1);
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* head.next = new ListNode(2);
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* head.next.next = new ListNode(3);
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* Solution solution = new Solution(head);
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*
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* // getRandom() should return either 1, 2, or 3 randomly. Each element should have
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* equal probability of returning.
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* solution.getRandom();
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***************************************************************************************/
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/**
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* Definition for singly-linked list.
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* struct ListNode {
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* int val;
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* ListNode *next;
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* ListNode(int x) : val(x), next(NULL) {}
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* };
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*/
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class Solution {
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public:
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/** @param head The linked list's head.
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Note that the head is guaranteed to be not null, so it contains at least one node. */
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Solution(ListNode* head) {
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this->head = head;
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this->len = 0;
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for(ListNode*p = head; p!=NULL; p=p->next, len++);
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srand(time(NULL));
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}
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/** Returns a random node's value. */
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int getRandom() {
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int pos = rand() % len;
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ListNode *p = head;
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for (; pos > 0; pos--, p=p->next);
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return p->val;
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}
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ListNode* head;
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int len;
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};
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/**
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* Your Solution object will be instantiated and called as such:
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* Solution obj = new Solution(head);
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* int param_1 = obj.getRandom();
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*/
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