Question

我正在编写一个代码来重塑信号。我正在通过不必要的重复获得输出。

INPUT：

String [] rani = {“A”，“1”，“2”，“OK”，“B”，“3”，“4”，“OK”，“B”，“1”，“ 3" ， “OK”};

必需的输出：

A / 3 B / 7 B / 4

获得了输出：

A / 3 A / 3 A / 3 A / 3 B / 7 B / 7 B / 7 B / 7 B / 4

算法：单字母串（“A”，“B”等）后跟数字串（“1”，“2”等）。每个字母表字符串后跟斜杠和数字的总和，字符串“OK”将被忽略。

作为java和编程的新手我需要帮助来获得所需的输出。

我的代码是：

public class SignalOK {

    public static void main(String[] arg) {
        String finalSignal = "";
        String netSignal = "";
        String name = "";
        int total = 0;
        String[] rani = { "A", "1", "2", "OK", "B", "3", "4", "OK", "B", "1",
                "3", "OK" };

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

            if ((rani[i] == "A") || (rani[i] == "B")) {

                name = rani[i];
            }

            if ((rani[i] == "1") || (rani[i] == "2") || (rani[i] == "3")
                    || (rani[i] == "4")) {

                total = total + Integer.valueOf(rani[i]);
            }

            if (rani[i] == "OK") {

                netSignal = name + "/" + String.valueOf(total) + "   ";
                name = "";
                total = 0;
            }
            finalSignal = finalSignal + netSignal;
        }

        System.out.println(finalSignal);
    }
}

Answer 1

如果括号：

，只需在"OK"内移动最终结果字符串连接

if (rani[i].equals("OK")) {
    netSignal = name + "/" + String.valueOf(total) + "   ";
    name = "";
    total = 0;   
    finalSignal = finalSignal + netSignal;
}

另外，请始终使用.equals()来比较字符串。

Answer 2

另一种方法： -

    public static void main(String[] args) {
    String[] inputString = {"A","1","2","OK","B","3","4","OK","B","1","3","OK"};
    StringBuilder sb = new StringBuilder();
    for(String s : inputString)
        sb.append(s); // creating a String from array contents
    String ss[] = sb.toString().split("OK"); // split on the basis of 'OK'
    sb.setLength(0); // emptying the sb, so that it can be used later on also
    for(String s : ss){
        char[] ch = s.toCharArray();
        sb.append(ch[0]); // first alphabet like 'A','B','C'
        int val = 0;
        for(int i = 1; i< ch.length ; i++)
            val +=Integer.parseInt(""+ch[i]); // calculate the int value
        sb.append("/"+val+" "); // finally appending alphabet with int value
    }
    System.out.println(sb); 
}

输出： - A/3 B/7 B/4

java循环输出中不需要的重复

2 个答案: