【leetcode】382. Linked List Random Node

一、题目描述

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();

方法一：用一个vector存储链表的每个元素，然后用random随机生成一个下标。

c++代码（60ms）

/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
private:
vector<int> element;

/** @param head The linked list's head.
Note that the head is guaranteed to be not null, so it contains at least one node. */
public:
Solution(ListNode* head) {
while(head!=NULL){
element.push_back(head->val);
head=head->next;
}//while
}

/** Returns a random node's value. */
int getRandom() {
int len=element.size();
return element[rand()%len];
}
};

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

方法二：有一个小技巧，可在遍历链表后就能马上得出结果。采用覆盖的形式，边遍历边计数已经到了第几个元素了，比如你在访问第2个元素时，count用来计数，此时count=2，如果rand()%count == 0，那么就存储当前节点的值到result中，那么命中的概率是多少呢，很显然rand()%count只能取0和1，所以概率是1/2，遍历还在继续。然后遍历第3个节点，如果第3个节点命中那么会把第3个节点的值存储到result中从而覆盖掉第2个节点的值。所以对于第2个节点，要想遍历完都不被覆盖（即选中了第2个节点）的概率这样算：1/2
* 2/3 * 3/4 * .... * (n-1)/n = 1/n。同理对于第x个节点，最后命中这个节点的概率这样算：(x-1)/x * x/(x+1) * (x+1)/(x+2) * (n-1)/n = 1/n 。所以所有的节点命中的概率都是1/n了。

c++代码（64ms）

/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
private:
ListNode* root;

/** @param head The linked list's head.
Note that the head is guaranteed to be not null, so it contains at least one node. */
public:
Solution(ListNode* head) {
root=head;
}

/** Returns a random node's value. */
int getRandom() {
int result=0;
ListNode *tmp=root;
for(int count=1;tmp!=NULL;count++,tmp=tmp->next){
if(rand()%count==0)
result=tmp->val;
}//for
return result;
}
};

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