leetcode 382. 链表随机节点(Linked List Random Node)

题目描述：

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();

---

解法：

/**
 * 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. */
    ListNode* head;
    Solution(ListNode* head) {
        this->head = head;
    }
    
    /** Returns a random node's value. */
    int getRandom() {
        
        ListNode* cur = head->next;
        int res = head->val;
        int n = 2;
        while(cur != NULL){
            int idx = rand()%n;
            if(idx == 0){
                res = cur->val;
            }
            cur = cur->next;
            n++;
        }
        return res;
    }
};

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