如何混合处理Java字符串

Question

Java字符串包含两种类型的单词。第一类词语是：

Wx1,Wx2,Wx3,Wx4,Wx5,Wx6, Wx7 etc.

第二类词语如下：

Wy1,Wy2 etc. 

给定的

String str= "Wx1 Wx2   Wy1   Wx3 Wx4   Wy2    Wx5 Wx7 Wx8"

第一类由function1()处理，第二类由function2()处理，function3()进一步处理function1()的级联输出。

function1(Wx1) gives String S1
function1(Wx2) gives String S2
String S12= S1 + S2;
    function3(S12) returns String str1


function2(Wy1) gives String Y1

function1(Wx3) gives String S3
function1(Wx4) gives String S4

String S34= S3 + S4;
    function3(S34) returns String str2

    function2(Wy2) gives String Y2

function1(Wx5) gives String S5
function1(Wx6) gives String S6
function1(Wx7) gives String S7
function1(Wx8) gives String S8

String S5678 = S5 + S6 + S7 + S8
    function3(S5678) returns String str3


String output = str1 + Y1+ str2 + Y2 + str3;

This is the whole logic my program.

在最后一步中，我对如何调用function3()的循环感到困惑：

for(int i=0; i<Words.length; i++){

if Word[i]= Type1??????????{ doThis(); ?????

//somewhere here I have to call function3(), how the program will judge that the next word is of type2, hence function3 is to be called here.// WHERE AND HOW TO CALL function3();
    }
else {doThat();
}

对于编程来说，我需要帮助来形成循环的逻辑。

Answer 1

我对字符串S12、S34等的多重创建感到困惑。最后，您的结果应该是S12 + Y1+ S34 + Y2 + S5678;。这相当于S1 + S2 + Y1 + S3 + S4 + Y2 + S5 + S6 + S7 + S8。那么，在这里创建这些中间级联有什么意义吗？如果没有，你可以试试这个程序。

编辑:

public class MixProcessor {
    private static int counter = 0;

    public static void main(String[] args) {
        String str= "Wx1 Wx2 Wy1 Wx3 Wx4 Wy2 Wx5 Wx7 Wx8";
        StringBuilder output = new StringBuilder();
        StringBuilder temp = new StringBuilder();
        String previousPatter = "";
        boolean shouldMerge = false;

        for(String s : str.split(" ")){
            String processedString = "";

            if (s.matches(".*?x.*")){
                shouldMerge = ("x".equals(previousPatter) || previousPatter.trim().length() == 0) ? true : false;
                previousPatter = "x";
                processedString = function1(s);
            } else if (s.matches(".*?y.*")){
                shouldMerge = ("y".equals(previousPatter) || previousPatter.trim().length() == 0) ? true : false;
                previousPatter = "y";
                processedString = function2(s);
            }

            if(shouldMerge){
                temp.append(processedString);
            } else {
                output.append(function3(temp.toString()));
                temp = new StringBuilder(processedString);
            }
        }
        //Append the last value
        output.append(function3(temp.toString()));

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

    private static String function2(String s) {
        return "Y" + s.substring(s.indexOf("y") + 1);
    }

    private static String function1(String s) {
        return "S" + s.substring(s.indexOf("x") + 1);
    }

    private static String function3(String s) {
        // Here this could be any function. I'm just trying to make as per the question
        return (s.indexOf("S") != -1) ? ("str" + (++counter)) : s;
    }

}