用java程序实现一道排列组合的数学题？

Question

0-9选6个数加起来等于33的排列组合,数字可以重复。

Answer 1

简单粗暴的排列算法：

javapublic class BruteForceArrangement {
  public static void main(String[] args) {
    for (int a = 0; a < 10; a++)
      for (int b = 0; b < 10; b++)
        for (int c = 0; c < 10; c++)
          for (int d = 0; d < 10; d++)
            for (int e = 0; e < 10; e++)
              for (int f = 0; f < 10; f++)
                if (a + b + c + d + e + f == 33)
                  System.out.printf("%d%d%d%d%d%d\n", a, b, c, d, e, f);
  }
}

简单粗暴的组合算法：

javapublic class BruteForceCombination {
  public static void main(String[] args) {
    for (int a = 0; a < 10; a++)
      for (int b = a; b < 10; b++)
        for (int c = b; c < 10; c++)
          for (int d = c; d < 10; d++)
            for (int e = d; e < 10; e++)
              for (int f = e; f < 10; f++)
                if (a + b + c + d + e + f == 33)
                  System.out.printf("%d%d%d%d%d%d\n", a, b, c, d, e, f);
  }
}

Answer 2

递归实现：

javapublic class Demo {
    public static void main(String[] args) {
        new Demo().go();
    }

    int[] numbers = new int[6];

    private final int result = 33;

    public void go() {
        compute(0, 0, 0);
    }

    private void compute(int index, int start, int n) {
        for (int i = start; i <= 10; i++) {
            numbers[index] = i;
            int m = n + i;

            if (m == result) {
                print();
            } else if (m > result) {
                break;
            } else if (index < 5) {
                compute(index + 1, i, m);
            }
        }
        numbers[index] = 0;
    }

    private void print() {
        StringBuffer buffer = new StringBuffer();
        for (int i = 0; i < 5; i++) {
            buffer.append(numbers[i])
                .append(" + ");
        }
        buffer.append(numbers[5])
            .append(" = ")
            .append(result);

        System.out.println(buffer.toString());
    }
}