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528. 按权重随机选择

给定一个正整数数组 w ，其中 w[i] 代表下标 i 的权重（下标从 0 开始），请写一个函数 pickIndex ，它可以随机地获取下标 i，选取下标 i 的概率与 w[i] 成正比。

例如，对于 w = [1, 3]，挑选下标 0 的概率为 1 / (1 + 3) = 0.25 （即，25%），而选取下标 1 的概率为 3 / (1 + 3) = 0.75（即，75%）。

也就是说，选取下标 i 的概率为 w[i] / sum(w) 。

 

示例 1：

输入：
["Solution","pickIndex"]
[[[1]],[]]
输出：
[null,0]
解释：
Solution solution = new Solution([1]);
solution.pickIndex(); // 返回 0，因为数组中只有一个元素，所以唯一的选择是返回下标 0。

示例 2：

输入：
["Solution","pickIndex","pickIndex","pickIndex","pickIndex","pickIndex"]
[[[1,3]],[],[],[],[],[]]
输出：
[null,1,1,1,1,0]
解释：
Solution solution = new Solution([1, 3]);
solution.pickIndex(); // 返回 1，返回下标 1，返回该下标概率为 3/4 。
solution.pickIndex(); // 返回 1
solution.pickIndex(); // 返回 1
solution.pickIndex(); // 返回 1
solution.pickIndex(); // 返回 0，返回下标 0，返回该下标概率为 1/4 。

由于这是一个随机问题，允许多个答案，因此下列输出都可以被认为是正确的:
[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]
......
诸若此类。

提示：

1 <= w.length <= 10000
1 <= w[i] <= 10^5
pickIndex 将被调用不超过 10000 次

528. Random Pick with Weight

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.

代码

var Solution = function(w) {
    pre = new Array(w.length).fill(0);
    pre[0] = w[0];
    for (let i = 1; i < w.length; ++i) {
        pre[i] = pre[i - 1] + w[i];
    }
    this.total = _.sum(w);
};

Solution.prototype.pickIndex = function() {
    const x = Math.floor((Math.random() * this.total)) + 1;
    const binarySearch = (x) => {
        let low = 0, high = pre.length - 1;
        while (low < high) {
            const mid = Math.floor((high - low) / 2) + low;
            if (pre[mid] < x) {
                low = mid + 1;
            } else {
                high = mid;
            }
        }
        return low;
    }
    return binarySearch(x);
};