382 Linked List Random Node 链表随机节点

Posted 2020-10-31 lina2014

给定一个单链表，随机选择链表的一个节点，并返回相应的节点值。保证每个节点被选的概率一样。
进阶:
如果链表十分大且长度未知，如何解决这个问题？你能否使用常数级空间复杂度实现？
示例:
// 初始化一个单链表 [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()方法应随机返回1,2,3中的一个，保证每个元素被返回的概率相等。
solution.getRandom();
详见：https://leetcode.com/problems/linked-list-random-node/description/

C++：

方法一：

/**
 * 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 guaranteed to be not null, so it contains at least one node. */
    Solution(ListNode* head) {
        len=0;
        this->head=head;
        ListNode *cur=head;
        while(cur)
        {
            ++len;
            cur=cur->next;
        }
    }
    
    /** Returns a random node\'s value. */
    int getRandom() {
        int t=rand()%len;
        ListNode *cur=head;
        while(t)
        {
            --t;
            cur=cur->next;
        }
        return cur->val;
    }
private:
    int len;
    ListNode *head;
};

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

 方法二：

/**
 * 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 guaranteed to be not null, so it contains at least one node. */
    Solution(ListNode* head) {
        this->head=head;
    }
    
    /** Returns a random node\'s value. */
    int getRandom() {
        int res=head->val;
        int i=2;
        ListNode *cur=head->next;
        while(cur)
        {
            int j=rand()%i;
            if(j==0)
            {
                res=cur->val;
            }
            ++i;
            cur=cur->next;
        }
        return res;
    }
private:
    ListNode *head;
};

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

 参考：https://www.cnblogs.com/grandyang/p/5759926.html