From Gossip@caterpillar

Algorithm Gossip: 洗撲克牌（亂數排列）

說明

洗撲克牌的原理其實與亂數排列是相同的，都是將一組數字（例如1～N）打亂重新排列，只不過洗撲克牌多了一個花色判斷的動作而已。

解法

初學者通常會直接想到，隨機產生1～N的亂數並將之存入陣列中，後來產生的亂數存入陣列前必須先檢查陣列中是否已有重複的數字，如果有這個數就不存入，再重新產生下一個數，運氣不好的話，重複的次數就會很多，程式的執行速度就很慢了，這不是一個好方法。

以1～52的亂數排列為例好了，可以將陣列先依序由1到52填入，然後使用一個迴圈走訪陣列，並隨機產生1～52的亂數，將產生的亂數當作索引取出陣列值，並與目前陣列走訪到的值相交換，如此就不用擔心亂數重複的問題了，陣列走訪完畢後，所有的數字也就重新排列了。

至於如何判斷花色？這只是除法的問題而已，取商數判斷花色，取餘數判斷數字，您可以直接看程式比較清楚。

實作：C Java Python Scala Ruby

#include <stdio.h>
#include <stdlib.h>
#include <time.h>
#define N 52

int main(void) {
    srand(time(0));
    int poker[N + 1];

    // 初始化陣列 
    int i;
    for(i = 1; i <= N; i++)
        poker[i] = i; 
    // 洗牌 
    for(i = 1; i <= N; i++) {
        int j = rand() % 52 + 1;
        int tmp = poker[i];
        poker[i] = poker[j]; 
        poker[j] = tmp; 
    }

    for(i = 1; i <= N; i++) {
        // 判斷花色 
        switch((poker[i]-1) / 13) { 
            case 0: printf("桃"); break;
            case 1: printf("心"); break;
            case 2: printf("磚"); break;
            case 3: printf("梅");
        } 

        // 撲克牌數字 
        int remain = poker[i] % 13;
        switch(remain) { 
            case 0: printf("K"); break;
            case 12: printf("Q"); break;
            case 11: printf("J"); break;
            default: printf("%d ", remain);
        } 
        printf("%c", i % 13 ? ' ' : '\n');
    } 

    return 0;
}

Java

public class ShuffleCard { 
    public static void main(String args[]) { 
        final int N = 52; 
        int[] poker = new int[N + 1]; 

        // 初始化陣列 
        for(int i = 1; i <= N; i++) 
            poker[i] = i; 

        // 洗牌 
        for(int i = 1; i <= N; i++) { 
            int j = (int) (Math.random() * N);

            if(j == 0)
                j = 1;

            int tmp = poker[i]; 
            poker[i] = poker[j]; 
            poker[j] = tmp; 
        } 

        for(int i = 1; i <= N; i++) { 
            // 判斷花色 
            switch((poker[i]-1) / 13) { 
                case 0: System.out.print("桃"); break; 
                case 1: System.out.print("心"); break; 
                case 2: System.out.print("磚"); break; 
                case 3: System.out.print("梅"); 
            } 

            // 撲克牌數字 
            int remain = poker[i] % 13; 
            switch(remain) { 
                case 0: System.out.print("K"); break; 
                case 12: System.out.print("Q"); break; 
                case 11: System.out.print("J"); break; 
                default: System.out.print(remain); 
            } 
            System.out.print(i % 13 == 0 ? '\n' : ' '); 
        } 
    } 
}

Python

import random
N = 52

poker = [0]
for i in range(1, N + 1):
    poker.append(i)
    
for i in range(1, N + 1):
    j = random.randint(1, N)
    poker[i], poker[j] = poker[j], poker[i]

card = {
          0: lambda : print("桃", end=""),
          1: lambda : print("心", end=""),
          2: lambda : print("磚", end=""),
          3: lambda : print("梅", end="")
       }

number = {
            0: lambda : print("K", end=" "),
           12: lambda : print("Q", end=" "),
           11: lambda : print("J", end=" ")
       }
       
for i in range(1, N + 1):
    card[(poker[i] - 1) // 13]()
    remain = poker[i] % 13
    number.get(remain, lambda : print(remain, end=" "))()
    print(end = "\n" if i % 13 == 0 else "")

Scala

import scala.util.Random
    
val N = 52
val poker = new Array[Int](N + 1)
for(i <- 1 to N) {
    poker(i) = i
}
val random = new Random
for(i <- 1 to N) {
    var j = (random.nextDouble * N).toInt
    if(j == 0) 
        j = 1
    val tmp = poker(i)
    poker(i) = poker(j)
    poker(j) = tmp
}

for(i <- 1 to N) {
    ((poker(i) - 1) / 13) match {
        case 0 => print("桃")
        case 1 => print("心")
        case 2 => print("磚")
        case 3 => print("梅")
    }    
    val remain = poker(i) % 13
    remain match {
        case 0 => print("K")
        case 12 => print("Q")
        case 11 => print("J")
        case _ => print(remain + " ")
    }
    print(if(i % 13 == 0) '\n' else ' ')
}

Ruby

# encoding: Big5
N = 52

poker = [0]

1.upto(N) { |i|
    poker << i
}

1.upto(N) { |i|
    j = (rand() * N).to_i
    poker[i], poker[j] = poker[j], poker[i]
}

1.upto(N) { |i|
    case (poker[i] - 1) / 13
    when 0
        print "桃"
    when 1
        print "心"
    when 2
        print "磚"
    when 3
        print "梅"
    end
    
    remain = poker[i] % 13
    case remain
    when 0
        print "K"
    when 12
        print "Q"
    when 11
        print "J"
    else
        print remain
    end
    print(i % 13 == 0 ? "\n" : " "); 
}