每日一题 382. 链表随机节点

382. 链表随机节点

简单

class Solution {
public:
    int  n;
    ListNode* head;
    Solution(ListNode* head) {
        int cnt = 0;
        this->head = head;
        while(head != 0)
        {
            head = head->next;
            ++cnt;
        }
        n =  cnt;
    }
    
    int getRandom() {
        int idx = rand() % n;
        ListNode* h = head;
        while(idx != 0)
        {
            h = h->next;
            idx--;
        }
        return h->val;

    }
};