Question

在我的代码中，我先分割一个字符串，然后将它们添加到这样的字符串中：

List<String> stringX = new ArrayList<String>();

但是在下一部分中，我需要检索字符串并将其转换为int []。我该如何处理？或者使用现有代码有更好的方法吗？

我曾尝试以几种不同的方式转换代码，但最终得到以下异常：ArrayIndexOutOfBoundsException和NumberFormatException。

List<String> stringX = new ArrayList<String>();
List<String> stringY = new ArrayList<String>();

public void convertString() {
    String input = getString(R.string.inputString);
    Pattern regex = Pattern.compile("\\((.*?)\\)");
    Matcher regexMatcher = regex.matcher(input);

    while (regexMatcher.find()) {
        list.add(regexMatcher.group(1));
    }
    for (String str : list) {
        String formatted = (str.replace(", ", ","));
        formattedList.add(formatted);
    }

    String[] values = formattedList.toArray(new String[formattedList.size()]);

    for (String s : values) {
        String coordXY[] = s.split(",");
        int x = Integer.parseInt(coordXY[0]);
        int y = Integer.parseInt(coordXY[1]);
        stringX.add(Integer.toString(x));
        stringY.add(Integer.toString(y));
    }
    difference();
}


private void difference(){
    // Setting the input array. These are what I need to get from the string lists
    int inputX[] = { 0, 1, 4, 4, 4, 0, 3, 2, 4 };
    int inputY[]= { 0, 3, 4, 2, 2, 1, 2, 3, 1 };

    // Obtaining the number of records
    int noOfRecordsX = inputX.length;
    int noOfRecordsY = inputY.length;

    // Array for storing differences
    int[] differenceX = new int [noOfRecordsX];
    int[] differenceY = new int [noOfRecordsY];

    differenceX [0] = 0; // First record difference is 0 only
    differenceY [0] = 0; // First record difference is 0 only

    // Looping number of records times
    for( int i=0; i < noOfRecordsX -1 ;i++)
    {
        // Difference = next record - current record
        differenceX [i+1]= inputX [i+1] - inputX[i];
    }

    for( int i=0; i < noOfRecordsY -1 ;i++)
    {
        // Difference = next record - current record
        differenceY [i+1]= inputY [i+1] - inputY[i];
    }
}

Answer 1

尝试一下：

private void testMethod(){
    String input = "(0, 0) (1, 3) (4, 4) (4, 2) (4, 2) (0, 1) (3, 2) (2, 3) (4, 1)";
    ArrayList<Integer> x = new ArrayList<>();
    ArrayList<Integer> y = new ArrayList<>();

    Pattern pattern = Pattern.compile("\\((.*?)\\)");
    Matcher matcher = pattern.matcher(input);

    while (matcher.find()) {
        String[] sa = matcher.group(1).split(",");
        x.add(Integer.valueOf(sa[0].trim()));
        y.add(Integer.valueOf(sa[1].trim()));
    }
}

希望您不介意我使用ArrayList<Integer>而不是int[]。

我确信有一些顶级RegEx专家可以在两行中完成所有这些工作，但这可以做到-而且很容易跟踪正在做的事情。

将字符串列表转换为Int []

1 个答案: