洗牌数组中去

问题描述

我试着翻译下面的Python code到去

I tried to translate the following Python code to Go

import random

list = [i for i in range(1, 25)]
random.shuffle(list)
print(list)

但发现我的围棋版本漫长而尴尬，因为没有洗牌的功能，我必须实现的接口，并转换类型。

but found my Go version lengthy and awkward because there is no shuffle function and I had to implement interfaces and convert types.

什么是我的code的惯用围棋版本？

What would be an idiomatic Go version of my code?

推荐答案

随着你的清单仅仅是个整数，从1到25，可以使用的：

As your list is just the integers from 1 to 25, you can use Perm :

list := rand.Perm(25)
for i, _ := range list {
    list[i]++
}

请注意，使用由 rand.Perm 给出一个排列洗牌任何阵列的有效途径。

Note that using a permutation given by rand.Perm is an effective way to shuffle any array.

dest := make([]int, len(src))
perm := rand.Perm(len(src))
for i, v := range perm {
    dest[v] = src[i]
}

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