数组列表<string>检查满足条件的第一个和最后一个值

Question

I have an arraylist of strings, called ArrayList<string> records that stores strings of data that will be delimited by commas , . At the [5] position of the string (after being delimited), will be a value between 0 to 500. An example of a string of that data is as follows:

userA, 1418600437, 38.9047, 177.0164, washington, 180

so anyway, the arraylist will have many strings of similar data. What I need is to iterate through the arraylist (which is already sorted in the sequence I want), then find the FIRST value in the arraylist that fulfills the criteria that the number at position[5] is < 200 and then the LAST value of the arraylist that fulfills the same criteria. When the first value is detected, a String will be assigned to that one record to call it the "start" value and the same for the last value. Can anyone help me with the logic, or just a pseudocode of how that might work?

Answer 1

Try the following code. The first value to match the criteria is recorded using a boolean flag. The last value is updated each time the criteria matches, with the result that after the for loop finishes, the variable lastValue will store the last matching value.

boolean seenFirst = false;
int firstValue, lastValue;

for (String record : records) {
    String[] parts = record.split(", ");
    int value = Integer.parseInt(parts[5]);

    if (value < 200)) {
        lastValue = value;
        if (!seenFirst) {
            seenFirst = true;
            firstValue = value;
        }
    }
}

Answer 2

If you are using Java 8, this is very simple:

List<string> records = new ArrayList<>();
String firstValue = records.stream().filter(s -> Integer.parseInt(s.split(",")[5].trim()) < 200)
                                    .findFirst().get();
String lastValue = IntStream.range(0, records.size())
                            .mapToObj(i -> records.get(records.size() - i - 1))
                            .filter(i -> Integer.parseInt(s.split(",")[5].trim()) < 200)
                            .findFirst().get();

To get the first value, we get the integer, filter those that are less than 200 and retrieve the first value.

To get the last value, it is the same but the iteration is done in reverse order.

Answer 3

As I stated in the comments: first do an iteration to get the first matching record, then do a reverse iteration to get the last matching record. In both cases, break out of your iterations after you found a match.

Could look something like this:

String first, last;

for (int i = 0; i < records.size(); i++) {
    if (Integer.parseInt(records[i].split(", ")[5]) < 200) {
        first = records[i];
        break;
    }
}

for (int i = records.size() - 1; i >=0; i--) {
    if (Integer.parseInt(records[i].split(", ")[5]) < 200) {
        last = records[i];
        break;
    }
}

Obviously should be refactored to get rid of the duplicated matching logic.