[LeetCode] 710. Random Pick with Blacklist
grandyang opened this issue · comments
Given a blacklist B
containing unique integers from [0, N)
, write a function to return a uniform random integer from [0, N)
which is NOT in B
.
Optimize it such that it minimizes the call to system’s Math.random()
.
Note:
1 <= N <= 1000000000
0 <= B.length < min(100000, N)
[0, N)
does NOT include N. See interval notation.
Example 1:
Input:
["Solution","pick","pick","pick"]
[[1,[]],[],[],[]]
Output: [null,0,0,0]
Example 2:
Input:
["Solution","pick","pick","pick"]
[[2,[]],[],[],[]]
Output: [null,1,1,1]
Example 3:
Input:
["Solution","pick","pick","pick"]
[[3,[1]],[],[],[]]
Output: [null,0,0,2]
Example 4:
Input:
["Solution","pick","pick","pick"]
[[4,[2]],[],[],[]]
Output: [null,1,3,1]
Explanation of Input Syntax:
The input is two lists: the subroutines called and their arguments. Solution
's constructor has two arguments, N
and the blacklist B
. pick
has no arguments. Arguments are always wrapped with a list, even if there aren't any.
这道题让我们生成一个N以内的随机数,但是还给了一个黑名单,意思是黑名单里面的数字不能被选到。于是博主最先想到的方法就是用拒绝采样Rejection Sampling来做,因为之前做过使用该方法的两道题 Implement Rand10() Using Rand7() 和 Generate Random Point in a Circle,所以可以立马想到。思路其实很简单,就是随机一个数,如果是黑名单里的,那么就重新随机。为了提高在黑名单中查找数字的速度,我们将所有黑名单的数字放到一个HashSet中,这样我们就拥有了常数级查找的速度,看似一切水到渠成,燃鹅被OJ强行打脸,TLE!那么换一种思路吧,既然你有黑名单,那么林北就有白名单,把所有没被block的数字都放到一个新数组中,然后随机生成数组坐标不就完了。燃鹅x2,又被OJ放倒了,MLE!不准用这么多内存。岂可修,真的没别的办法了嘛?!还好方法解答贴中给了一种使用HashMap的方法来做,博主仔细研读了一番,发现确实秒啊!既然数字总共有N个,那么减去黑名单中数字的个数,就是最多能随机出来的个数。比如N=5,黑名单中有两个数{2, 4},那么我们最多只能随机出三个,但是我们如果直接rand()%3,会得到0,1,2,我们发现有两个问题,一是黑名单中的2可以随机到,二是数字3没法随机到。那么我们想,能不能随机到0或1则返回其本身,而当随机到2到时候,我们返回的是3,我们需要建立这样的映射,这就是使用HashMap的动机啦。我们首先将超过N - blacklist.size()的数字放入一个HashSet,这里就把{3, 4}放进去了,然后我们遍历blacklist中的数字,如果在HashSet中的话,就将其删除,这样HashSet中就只有{3}了,这个需要建立映射的数字,而用什么数字建立,当然是用黑名单中的数字了,遍历黑名单中的数字,如果小于N - blacklist.size()的话,说明是有可能随机到的,我们和HashSet中的第一个数字建立映射,然后我们可以用个iterator,指向HashSet中的下一个数组,然后继续建立映射。从而实现在pick函数中的移魂换影大法了,先随机个数字,如果有映射,则返回映射值,否则返回原数字,参见代码如下:
class Solution {
public:
Solution(int N, vector<int> blacklist) {
unordered_set<int> st;
len = N - blacklist.size();
for (int i = len; i < N; ++i) st.insert(i);
for (int num : blacklist) st.erase(num);
auto it = st.begin();
for (int num : blacklist) {
if (num < len) m[num] = *it++;
}
}
int pick() {
int k = rand() % len;
return m.count(k) ? m[k] : k;
}
private:
unordered_map<int, int> m;
int len;
};
类似题目:
参考资料:
https://leetcode.com/problems/random-pick-with-blacklist/