<LeetCode OJ> 382. Linked List Random Node

标签：

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();

分析：

给一个链表，随机返回一个节点，那么最直接的方法就是先统计出链表的长度，然后根据长度随机生成一个位置，然后从开头遍历到这个位置即可。这里考虑到他的调用方式，所以先用vector数组存储的链表中的每一个值，随机寻找值时不再重新遍历链表！参见代码如下：

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    /** @param head The linked list's head. Note that the head is guanranteed to be not null, so it contains at least one node. */
    Solution(ListNode* head) {
        m_nlen=0;
        ListNode* move=head;
        while(move!=NULL)
        {
            m_nlen++;
            result.push_back(move->val);
            move=move->next;
        }
    }
    
    /** Returns a random node's value. */
    int getRandom() {
        return result[rand()%m_nlen];
    }
private:
    int m_nlen;
     vector<int> result;
    
};

/**
 * Your Solution object will be instantiated and called as such:
 * Solution obj = new Solution(head);
 * int param_1 = obj.getRandom();
 */

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原文地址：http://blog.csdn.net/ebowtang/article/details/52187799

原作者博客：http://blog.csdn.net/ebowtang

本博客LeetCode题解索引：http://blog.csdn.net/ebowtang/article/details/50668895

<LeetCode OJ> 382. Linked List Random Node

标签：

原文地址：http://blog.csdn.net/ebowtang/article/details/52187799