更簡單的方法來做很多嵌套for循環

Question

我需要生成所有可能的顏色組合（我有8種不同的顏色）的列表，並且希望指定輸出的length，而無需進入代碼並添加另一個循環。

for (int s1 = 0; s1 < kinds.length; s1++) { 
     for (int s2 = 0; s2 < kinds.length; s2++) { 
      for (int s3 = 0; s3 < kinds.length; s3++) { 
       for (int s4 = 0; s4 < kinds.length; s4++) { 
        for (int s5 = 0; s5 < kinds.length; s5++) { 
         String[] guess = new String[length]; 
         guess[0] = colors[s1]; 
         guess[1] = colors[s2]; 
         guess[2] = colors[s3]; 
         guess[3] = colors[s4]; 
         guess[4] = colors[s5]; 
         Possible.add(guess); 

        } 
       } 
      } 
     } 
    } 

Answer 1

您可以使用遞歸而不是循環。總的想法是所有可能的組合都可以從所有可能的第一顏色concat與其餘的組合中產生。

private static List<String> possibles = new ArrayList<>(); 

private static void combi(char[] arr, char[] out, int start, int end, int index, int length) { 
    if (index == length) { 
     possibles.add(new String(out)); 
     return; 
    } 

    for (int i = start; i <= end && end - i + 1 >= length - index; i++) { 
     out[index] = arr[i]; 
     combi(arr, out, i + 1, end, index + 1, length); 
    } 
} 

private static void compute(char[] colors, int length) { 
    char[] output = new char[length]; 
    combi(colors, output, 0, colors.length - 1, 0, length); 
} 

public static void main(String[] args) { 
    char[] colors = {'a', 'b', 'c', 'd', 'e'}; 
    int length = 3; 
    compute(colors, length); 

    for (String possible : possibles) { 
     System.out.println(possible); 
    } 
}