Question

我正在尝试使用类似于以下的字符串：3A5o2n4t并将其扩展回这样的字符串：AAAooooonntttt（字母前面的数字是字母重复的次数）我试图使用Integer.parseInt（）来获取字母前面的数字，但它会捕获所有数字。有没有办法一次抓一个号码？另外，解决该问题后，我的代码看起来还可以吗？还是我还想念一点？

public String runLengthDecoding(String str3) {
          String convert = "";
          int number = 0;
          if (! str3.isEmpty()) {
             convert = str3.charAt(0) + "";
          }
          for (int i = 0; i <= str3.length() - 1; i++) { 
             if (Character.isDigit(str3.charAt(i))) { //true or false, the current character is a digit
                String temp = "" + str3.charAt(i); //if true, make that character a string
                number = Integer.parseInt(temp); /*now store that character as a number (but I only want the current
                                                   number, not all the numbers in the string*/
                System.out.println(number); /*Testing to see what number is, which is where I found it was 
                                               storing all the numbers */
                String temp2 = str3.charAt(i + 1) + ""; //Its supposed to start making a string that prints the character in front of it
                convert = temp2.repeat(number); //it will print the character however many times that number was stored as
             }

       }
       return convert;
       }

我还没有学会如何使用数组，这就是为什么我不使用数组。

Answer 1

编辑为：
-容纳长度大于1的字符串。示例：10AA
-容纳以字符串开头的输入。例如：A5o

要解决此问题，您必须获取所有同时出现的数字，例如，如果您具有“ 55s”，则必须获取“ 55”，这就是您的代码不正确的原因，因为如果您每当看到数字就解析parInt，它将立即解析为5，但实际数字为55，因此，您应该先累积并发数字，并且在遇到第一个非数字时只累积parseInt。

有关详细信息，请参见代码和注释：

public class Main {
    public static void main(String[] args) {
        System.out.println("Input: 3A5o2n4t => Output : " + runLengthDecoding("3A5o2n4t"));
        System.out.println("Input: 3AA5o2n4t => Output : " + runLengthDecoding("3AA5o2n4t"));
        System.out.println("Input: 10A5o2n4t => Output : " + runLengthDecoding("10A5o2n4t"));
        System.out.println("Input: 10AA5o2n4t => Output : " + runLengthDecoding("10AA5o2n4t"));
        System.out.println("Input: A5o => Output : " + runLengthDecoding("A5o"));
        System.out.println("Input: AB5o => Output : " + runLengthDecoding("AB5o"));
    }

    public static String runLengthDecoding(String str3) {
        String convert = "";
        int number = 0;
        String numberString = "";
        String toBeRepeatedString = "";
        boolean flag = false;
        for (int i = 0; i <= str3.length() - 1; i++) {
            char currentChar = str3.charAt(i);
            if (Character.isDigit(currentChar)) { // true or false, the current character is a digit
                numberString = numberString + currentChar; // store the possible integer
            } else {
                if (i + 1 < str3.length()) {
                    char nextChar = str3.charAt(i + 1); // check if the next char is a digit
                    if (!Character.isDigit(nextChar)) { // if not a digit then append toBeRepeatedString

                        if (i == 0 || i + 1 >= str3.length()) {
                            flag = true;
                        } else {
                            toBeRepeatedString += nextChar;
                            flag = false;
                        }
                    } else {
                        flag = true;
                    }
                }

                if (flag) {
                    toBeRepeatedString += currentChar;

                    // This will accomodate inputs "A3B";
                    if (!numberString.isEmpty()) {
                        number = Integer.parseInt(numberString); // parse the number of repeats
                    } else {
                        number = 1;
                    }

                    numberString = ""; // reset number

                    String temp2 = "";

                    // Repeat the currentChar
                    for (int j = 0; j < number; j++) {
                        temp2 += toBeRepeatedString;
                    }

                    convert = convert + temp2; // store it to the result
                    toBeRepeatedString = ""; // reset toBeRepeatedString
                }

            }

        }
        return convert;
    }

}

结果：

Input: 3A5o2n4t => Output : AAAooooonntttt
Input: 3AA5o2n4t => Output : AAAAAAooooonntttt
Input: 10A5o2n4t => Output : AAAAAAAAAAooooonntttt
Input: 10AA5o2n4t => Output : AAAAAAAAAAAAAAAAAAAAooooonntttt
Input: A5o => Output : Aooooo
Input: AB5o => Output : ABooooo

Answer 2

这是解决上述问题的最佳方法，它将处理您所有的情况：

public static void main(String[] args) {
    String input = "5a2s3T66e";
    System.out.println("Input is: "+input+" and output is: "+expandingCondenseString(input));
}
private static String expandingCondenseString(String input){
    StringBuilder result = new StringBuilder();
    String size = "";
    String value = "";
    for (int i=0;i<input.length();i++){
        if (Character.isDigit(input.charAt(i))) {
            size = size + input.charAt(i);
        } else {
            value = value + input.charAt(i);
            if(i+1<input.length() && !Character.isDigit(input.charAt(i+1))){
                continue;
            }
            if(size.isEmpty()){
                size = "1";
            }
            for (int j=0;j<Integer.parseInt(size);j++){
                result.append(value);
            }
            size = "";
            value = "";
        }
    }
    return String.valueOf(result);
}

扩展压缩字符串

2 个答案: