剑指 Offer II 071. 按权重生成随机数

题目描述

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：
inputs = ["Solution","pickIndex"]
inputs = [[[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：
inputs = ["Solution","pickIndex","pickIndex","pickIndex","pickIndex","pickIndex"]
inputs = [[[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

注意：本题与主站 528 题相同： https://leetcode.cn/problems/random-pick-with-weight/

解法

方法一

Python3JavaC++Go

class Solution:
    def __init__(self, w: List[int]):
        n = len(w)
        self.presum = [0] * (n + 1)
        for i in range(n):
            self.presum[i + 1] = self.presum[i] + w[i]

    def pickIndex(self) -> int:
        n = len(self.presum)
        x = random.randint(1, self.presum[-1])
        left, right = 0, n - 2
        while left < right:
            mid = (left + right) >> 1
            if self.presum[mid + 1] >= x:
                right = mid
            else:
                left = mid + 1
        return left


# Your Solution object will be instantiated and called as such:
# obj = Solution(w)
# param_1 = obj.pickIndex()

class Solution {
    private int[] presum;

    public Solution(int[] w) {
        int n = w.length;
        presum = new int[n + 1];
        for (int i = 0; i < n; ++i) {
            presum[i + 1] = presum[i] + w[i];
        }
    }

    public int pickIndex() {
        int n = presum.length;
        int x = (int) (Math.random() * presum[n - 1]) + 1;
        int left = 0, right = n - 2;
        while (left < right) {
            int mid = (left + right) >> 1;
            if (presum[mid + 1] >= x) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }
        return left;
    }
}

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

class Solution {
public:
    vector<int> presum;

    Solution(vector<int>& w) {
        int n = w.size();
        presum.resize(n + 1);
        for (int i = 0; i < n; ++i) presum[i + 1] = presum[i] + w[i];
    }

    int pickIndex() {
        int n = presum.size();
        int x = rand() % presum[n - 1] + 1;
        int left = 0, right = n - 2;
        while (left < right) {
            int mid = left + right >> 1;
            if (presum[mid + 1] >= x)
                right = mid;
            else
                left = mid + 1;
        }
        return left;
    }
};

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

type Solution struct {
    presum []int
}

func Constructor(w []int) Solution {
    n := len(w)
    pre := make([]int, n+1)
    for i := 0; i < n; i++ {
        pre[i+1] = pre[i] + w[i]
    }
    return Solution{pre}
}

func (this *Solution) PickIndex() int {
    n := len(this.presum)
    x := rand.Intn(this.presum[n-1]) + 1
    left, right := 0, n-2
    for left < right {
        mid := (left + right) >> 1
        if this.presum[mid+1] >= x {
            right = mid
        } else {
            left = mid + 1
        }
    }
    return left
}

/**
 * Your Solution object will be instantiated and called as such:
 * obj := Constructor(w);
 * param_1 := obj.PickIndex();
 */

剑指 Offer II 071. 按权重生成随机数

题目描述

解法

方法一

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