获得的字符的所有可能组合以阵列 [英] Get all possible combinations of characters in an array

问题描述

我有字符C [] []具有不同映射的每个索引的阵列。
例如：

I have an array of characters c[][] with different mappings to each index. For example:

{'a', 'b', 'c', 'd', 'e', 'f' } {'g', 'h', 'i' }

我需要返回所有可能的字符组合为这个数组作为字符串。
那意思，对于上面的字符数组，我应该返回：
股份公司，啊，人工智能，BG，BH，比，CG，CH，CI等。
这将是很容易的只有两个像上面的东西字符数组做到这一点，但如果有多个阵列，那么我不知道该怎么办......
这就是我要求你们帮我！ ：）

I need to return all the possible character combinations for this array as a string. That meaning, for the above character array, I should return: "ag", "ah", "ai", "bg", "bh", "bi", "cg", "ch", "ci", etc. It would be easy to do this for a character array of only two things like above, but if there are more arrays, then I do not know what to do... Which is what I am asking you all to help me with! :)

推荐答案

有关两个数组的嵌套循环应该做的：

For two arrays two nested loops should do:

for (int i = 0 ; i != c[0].length ; i++) {
    for (int j = 0 ; j != c[1].length ; j++) {
        System.out.writeln(""+c[0][i]+c[1][j]);
    }
}

有关更多的嵌套，你需要一个递归或等效的基于堆栈的解决方案。

For more nesting you would need a recursive or an equivalent stack-based solution.

void combos(int pos, char[][] c, String soFar) {
    if (pos == c.length) {
         System.out.writeln(soFar);
         return;
    }
    for (int i = 0 ; i != c[pos].length ; i++) {
        combos(pos+1, c, soFar + c[pos][i]);
    }
}

从调用该递归函数的的main（）是这样的：

combos(0, c, "");

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