Description
You are given a 0-indexed array of positive integers w
where w[i]
describes the weight of the ith
index.
You need to implement the function pickIndex()
, which randomly picks an index in the range [0, w.length - 1]
(inclusive) and returns it. The probability of picking an index i
is w[i] / sum(w)
.
- For example, if
w = [1, 3]
, the probability of picking index0
is1 / (1 + 3) = 0.25
(i.e.,25%
), and the probability of picking index1
is3 / (1 + 3) = 0.75
(i.e.,75%
).
Example 1:
Input [“Solution”,“pickIndex”] [1,[]] Output [null,0]
Explanation Solution solution = new Solution([1]); solution.pickIndex(); // return 0. The only option is to return 0 since there is only one element in w.
Example 2:
Input [“Solution”,“pickIndex”,“pickIndex”,“pickIndex”,“pickIndex”,“pickIndex”] [1,3,[],[],[],[],[]] Output [null,1,1,1,1,0]
Explanation Solution solution = new Solution([1, 3]); solution.pickIndex(); // return 1. It is returning the second element (index = 1) that has a probability of 3/4. solution.pickIndex(); // return 1 solution.pickIndex(); // return 1 solution.pickIndex(); // return 1 solution.pickIndex(); // return 0. It is returning the first element (index = 0) that has a probability of 1/4.
Since this is a randomization problem, multiple answers are allowed. All of the following outputs can be considered correct: [null,1,1,1,1,0] [null,1,1,1,1,1] [null,1,1,1,0,0] [null,1,1,1,0,1] [null,1,0,1,0,0] … and so on.
Constraints:
1 <= w.length <= 104
1 <= w[i] <= 105
pickIndex
will be called at most104
times.
Code
因為 rand() % n
代表在 0 ~ n - 1
當中隨機選一個數,且題目的限制 1 <= w[i] <= 105
,所以 int random_number = rand() % w_.back() + 1;
(要加一,產生 1 ~ n
)。
以 w = [1, 3, 4]
為例,prefix sum 為 [1, 4, 8]
,random 出來的數字落在1
回傳的結果是 1
,落在 2,3,4
回傳的結果都是 2
,落在 5,6,7,8
的結果都是 3
,所以我們使用 lower_bound
而非 upper_bound
。