有没有什么算法来实现输出这个组合？ 输入：

  ARR1 = {X，Y，Z}
ARR2 = {A，B}
 

输出：

  XA，雅，ZA
XA，雅，ZB
XA，YB，ZA
XA，YB，ZB
XB，雅，ZA
XB，雅，ZB
XB，YB，ZA
XB，YB，ZB
 

解决方案

 静态的char [] ARR1 = {'X'，'Y'，'Z'};
静态的char [] ARR2 = {'A​​'，'B'};

公共静态无效的主要（字串[] args）{
    打印（新的char [arr1.length  -  1]，0）;
}

静态无效打印（的char []店，诠释深度）{
    对于（字符C：ARR2）{
        如果（深度LT; store.length）{
            店[深度] = C;
            打印（存储，深度+ 1）;
        } 其他 {
            的for（int i = 0; I＆LT; store.length;我++）{
                System.out.print（ARR1 [I] ++商店[I] +，）;
            }
            的System.out.println（ARR1 [深度] ++ C）;
        }
    }
}
 

编辑：忍不住尝试@ DenisKulagin的方法，所以这里有云：

 公共静态无效的主要（字串[] args）{
    的char [] ARR1 = {'X'，'Y'，'Z'};
    的char [] ARR2 = {'A​​'，'B'};

    对（INT I = 0; I＆。1所述;＆所述; arr1.length;我++）{
        对于（INT J = 0; J＆LT; arr1.length; J ++）{
            INT倒= arr1.length  -  1  -  J;
            INT指数=（ⅰ及（1＆其中;＆其中;倒））＆GT;＆GT;＆GT;倒;
            System.out.print（ARR1 [J] +，+ ARR2 [指数] +，）;
        }
        的System.out.println（）;
    }
}
 

没有这么灵活，我的版本，因为 ARR2 只能包含两个元素，但绝对是一个聪明的做法。

Is there any algorithm to achieve this combination of output? Input :

arr1 = {x, y, z}
arr2 = {a, b}

Output :

xa, ya, za
xa, ya, zb
xa, yb, za
xa, yb, zb
xb, ya, za
xb, ya, zb
xb, yb, za
xb, yb, zb

static char[] arr1 = {'x', 'y', 'z'};
static char[] arr2 = {'a', 'b'};

public static void main(String[] args) {
    print(new char[arr1.length - 1], 0);
}

static void print(char[] store, int depth) {
    for(char c : arr2) {
        if(depth < store.length) {
            store[depth] = c;
            print(store, depth + 1);
        } else {
            for(int i = 0; i < store.length; i++) {
                System.out.print(arr1[i] + "" + store[i] + ", ");
            }
            System.out.println(arr1[depth] + "" + c);
        }
    }
}

EDIT: Couldn't resist trying out @DenisKulagin's method, so here goes:

public static void main(String[] args) {
    char[] arr1 = {'x', 'y', 'z'};
    char[] arr2 = {'a', 'b'};

    for(int i = 0; i < 1 << arr1.length; i++) {
        for(int j = 0; j < arr1.length; j++) {
            int inverted = arr1.length - 1 - j;
            int index = (i & (1 << inverted)) >>> inverted;
            System.out.print(arr1[j] + "" + arr2[index] + " ");
        }
        System.out.println();
    }
}

Not quite as flexible as my version, since arr2 can only contain 2 elements, but definitely a clever approach.