382. 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:

// 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

最简单的思路是构造的时候统计一下链表长度，每次生成随机数往后走就行了。

代码

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
    ListNode *pCur, *pHead;
    int step = 10;
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) {
        pCur = head;
        pHead = head;
        step = 0;
        ListNode *pTemp = head;
        while(pTemp != NULL){
            step++;
            pTemp = pTemp->next;
        }
    }
    
    /** Returns a random node's value. */
    int getRandom() {
        int step_cur = rand()%step;
        for(int i = 0; i < step_cur; i++){
            if(pCur->next == NULL)
                pCur = pHead;
            else
                pCur = pCur->next;
        }
        return pCur->val;
    }
};

感想

居然这就通过了。。。