Description

You are given an array of positive integers w where w[i] describes the weight of ith index (0-indexed).

We need to call the function pickIndex() which randomly returns an integer in the range [0, w.length - 1]. pickIndex() should return the integer proportional to its weight in the w array. For example, for w = [1, 3], the probability of picking the index 0 is 1 / (1 + 3) = 0.25 (i.e 25%) while the probability of picking the index 1 is 3 / (1 + 3) = 0.75 (i.e 75%).

More formally, the probability of picking index i is w[i] / sum(w).

Example 1:

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

Explanation
Solution solution = new Solution([1]);
solution.pickIndex(); // return 0. Since there is only one single element on the array the only option is to return the first element.

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's returning the second element (index = 1) that has probability of 3/4.
solution.pickIndex(); // return 1
solution.pickIndex(); // return 1
solution.pickIndex(); // return 1
solution.pickIndex(); // return 0. It's returning the first element (index = 0) that has probability of 1/4.

Since this is a randomization problem, multiple answers are allowed so 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 <= 10000
  • 1 <= w[i] <= 10^5
  • pickIndex will be called at most 10000 times.

分析

题目的意思是：实现一个带权重的随机取数，使得每个位置以给定权重取出。这里有一个trick，就是权重求和，比如[1,3]，求和后[1,4]，然后随机生成一个[0,3]的整数，落在1以前就是1，落在[1,3]上就是3的位置了哈。如果能搞懂这个剩下的就是生成的随机正数落在哪个区间的问题了。

代码

class Solution:

    def __init__(self, w: List[int]):
        self.w=[w[0]]
        for i in range(1,len(w)):
            self.w.append(self.w[-1]+w[i])
            
    def pickIndex(self) -> int:
        l=0
        h=len(self.w)-1
        x=random.randint(0,self.w[-1]-1)
        while(l<h):
            mid=l+(h-l)//2
            if(self.w[mid]<=x):
                l=mid+1
            else:
                h=mid
        return l
                
# Your Solution object will be instantiated and called as such:
# obj = Solution(w)
# param_1 = obj.pickIndex()

参考文献

​​[LeetCode] A simply readable Python​​

​[LeetCode] Random Pick with Weight 根据权重随机取点