多个字符串相乘

Question

I came across this question :

Given

{1, 2, 3}{a, b} 

convert into

1a, 1b, 2a, 2b, 3a, 3b

There can be any number of curly braces. Input will be in the form of List> or a multidimensional array. I can only think of having multiple for loops and multiplying the strings with each other.

Can someone help me with question in a better way?

I have tried having multiple loops for the each string with {} but that will not a good solution in case we have more number of curly braces

Answer 1

What about this.

• First you convert the input ( "{1,2,3}{a,b}{X,Y}" ) to "1,2,3}{a,b}{X,Y"
• Then you convert this to an String[] : ["1,2,3","a,b","X,Y"]
• Finally, you convert this to a List : {["1","2","3"],["a","b"],["X","Y"]}

What you do next is that you multiply the first two elements and add the result of the multiplication to the end of the list. You continue doing this, until there is just one element left: the final result. So:

• {["1","2","3"],["a","b"],["X","Y"]} becomes {["X", "Y"],["1a", "1b", "2a", "2b", "3a", "3c"]}

• {["X", "Y"],["1a", "1b", "2a", "2b", "3a", "3c"]} becomes {["X1a", "X1b", "X2a", "X2b", "X3a", "X3b", "Y1a", "Y1b", "Y2a", "Y2b", "Y3a", "Y3b"]}

   public static void main(String[] args) { String input = "{1,2,3}{a,b}{X,Y}"; System.out.println(multiplyAll(input)); } public static String multiplyAll(String input) { String[] parts = input.substring(1, input.length() - 1).split("\\\\}\\\\{"); List<String[]> elements = new ArrayList<>(); for (String part: parts) { elements.add(part.split(",")); } while (elements.size() > 1) { String[] parts1 = elements.remove(0); String[] parts2 = elements.remove(0); elements.add(multiply(parts1, parts2)); } String result = ""; for (String element: elements.get(0)) { result += element + ","; } return result.substring(0, result.length() - 1); } public static String[] multiply(String[] parts1, String[] parts2) { String[] result = new String[parts1.length * parts2.length]; for (int i = 0; i < parts1.length; i++) { for (int j = 0; j < parts2.length; j++) { result[i * parts2.length + j] = parts1[i] + parts2[j]; } } return result; }