470. 用 Rand7() 实现 Rand10() 已有方法 rand7 可生成 1 到 7 范围内的均匀随机整数,试写一个方法 rand10 生成 1 到 10 范围内的均匀随机整数. 不要使用系统的 Math.random() 方法. 示例 1: 输入: 1 输出: [7] 示例 2: 输入: 2 输出: [8,4] 示例 3: 输入: 3 输出: [8,1,10] 提示: rand7 已定义. 传入参数: n 表示 rand10 的调用次数. 进阶: rand7()调用次数的 期望值 是…
题目描述 已有方法 rand7 可生成 1 到 7 范围内的均匀随机整数,试写一个方法 rand10 生成 1 到 10 范围内的均匀随机整数. 不要使用系统的 Math.random() 方法. 示例 1: 输入: 1 输出: [7] 示例 2: 输入: 2 输出: [8,4] 示例 3: 输入: 3 输出: [8,1,10] 提示: rand7 已定义. 传入参数: n 表示 rand10 的调用次数. 进阶: rand7()调用次数的 期望值 是多少 ? 你能否尽量少调用 rand7() …
题目链接 https://leetcode-cn.com/problems/implement-rand10-using-rand7/ 题意: 给定一个rand7()的生成器,求解如何产生一个rand10()的生成器.注意这里定义rand7()的范围是17,而不是06 思路: 关键在于产生的结果必须是等概率的,这是一类非常泛化的题.具体而言涉及到由任意的randm()转化到randn()的计算方式. 详细解读可以参考:https://blog.csdn.net/u010025211/articl…
已有方法 rand7 可生成 1 到 7 范围内的均匀随机整数,试写一个方法 rand10 生成 1 到 10 范围内的均匀随机整数. public class Solution { public static int rand10() { int ans = rand2(); for (int i = 0; i < 10; i++) { ans = ans << 1; //二进制左移一位 ans = ans ^ rand2();//将随机生成的0或1,添加到二级制尾部 } return…
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