【LeetCode】 528. Random Pick with Weight

Description

Given an array w of positive integers, where w[i] describes the weight of index i, write a function pickIndex which randomly picks an index in proportion to its weight.

Note:

1. 1 <= w.length <= 10000
2. 1 <= w[i] <= 10^5
3. pickIndex will be called at most 10000 times.

Explanation of Input Syntax:
The input is two lists: the subroutines called and their arguments. Solution's constructor has one argument, the array w. pickIndex has no arguments. Arguments are always wrapped with a list, even if there aren't any.

給予一正整數陣列w，其中w[i]描述索引為i的權重。寫一個函式pickIndex會按照權重去隨機挑選一個索引。

注意：

1. 1 <= w.length <= 10000
2. 1 <= w[i] <= 10^5
3. pickIndex 至少會被呼叫 10000 次。

輸入語法解釋：
輸入會是兩個列表：哪些副程式被呼叫和它們的參數。
Solution的建構子有一個參數：陣列w。pickIndex沒有參數。參數總是被包裝在列表中，即便沒有任何參數。

Example:

Example 1:

Input: 
["Solution","pickIndex"]
[[[1]],[]]
Output: [null,0]


Example 2:

Input: 
["Solution","pickIndex","pickIndex","pickIndex","pickIndex","pickIndex"]
[[[1,3]],[],[],[],[],[]]
Output: [null,0,1,1,1,0]

Solution

• 使用rand()可以得到一個隨機值，介於0到int的最大值。
• 給予權重時，使用另一個vector去儲存它；並將元素改為遞增。
  • 如果權重為[1, 2, 3]，新的vector為[1, 3, 6]。
• 接著我們將rand() % sum(w)，可以得到一個0 ~ sum(w)的數字。
• 最後去找這個隨機數落在vector的哪個區間，就代表選到哪個數字。
  • 權重越大的索引，對應到的區間也越大，所以越容易被選中，藉此達到權重隨機。
• 下方code中，search的部分使用線性搜尋，可改用二分搜尋來加速。

Code

class Solution {
public:
    Solution(vector<int>& w) {
        int sum = 0;
        for(int i = 0; i < w.size(); i++)
        {
            sum += w[i];
            this->w.push_back(sum);
        }
    }
    
    int pickIndex() {
        int r = rand() % w.back();
        for(int i = 0; i < w.size(); i++)
            if(r < w[i])
                return i;
        return w.size() - 1;
    }
private:
    vector<int> w;
};

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

tags: LeetCode C++