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：DONE
給一個鏈表，隨機返回一個節點，那麼最直接的方法就是先統計出鏈表的長度，然後根據長度隨機生成一個位置，然後從開頭遍歷到這個位置即可。這裏考慮到他的調用方式，所以先用vector數組存儲的鏈表中的每一個值，隨機尋找值時不再重新遍歷鏈表！參見代碼如下：
/**
* 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();
*/注：本博文爲EbowTang原創，後續可能繼續更新本文。如果轉載，請務必複製本條信息！
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題解索引：http://blog.csdn.net/ebowtang/article/details/50668895