Java：将按字符串长度排序的字符串数组按字符串长度拆分为多个数组

Question

我目前有一个按字符串长度排序的字符串数组，例如：

String[] array = [a,b,c,ab,cd,abc,abcde,fghij,klmno]

如何根据字符串大小将此数组转换为多个数组，同时跟踪每个数组的字符串大小？我想要的是：

String[] array1 = [a,b,c]
String[] array2 = [ab,cd]
String[] array3 = [abc]
String[] array5 = [abcde,fghij,klmno]

我可能正在考虑为此使用矩阵，但不知道要这样做。

Answer 1

使用System:arraycopy

仅使用数组的解决方案：

import java.util.Arrays;

public class Main {
    public static void main(String[] args) {
        String[][] arraysList = new String[1][];
        String[] array = { "a", "b", "c", "ab", "cd", "abc", "abcde", "fghij", "klmno" };
        int srcPos, row = 0;
        for (int i = 0; i < array.length; i++) {
            srcPos = i;
            while (i < array.length - 1 && array[i].length() == array[i + 1].length()) {
                i++;
            }
            // Create a new array to store the current set of strings of equal length
            String[] subarray = new String[i - srcPos + 1];

            // Copy the current set of strings of equal length from array to subarray[]
            System.arraycopy(array, srcPos, subarray, 0, subarray.length);

            // Assign subarray[] to arraysList[][]
            arraysList[row++] = subarray;

            // Copy arraysList[][] to temp [][], increase size of arraysList[][] and restore
            // arrays from temp [][] to arraysList[][]
            String[][] temp = arraysList;
            arraysList = new String[row + 1][subarray.length];
            for (int j = 0; j < temp.length; j++) {
                arraysList[j] = temp[j];
            }
        }

        // Drop the last row which was created to store a new subarray but there was no
        // more subarrays to store and therefore it is empty.
        arraysList = Arrays.copyOf(arraysList, arraysList.length - 1);

        // Display the subarrays
        for (String[] arr : arraysList) {
            System.out.println(Arrays.toString(arr));
        }
    }
}

输出：

[a, b, c]
[ab, cd]
[abc]
[abcde, fghij, klmno]

使用List 和数组的解决方案：

import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;

public class Main {
    public static void main(String[] args) {
        List<String[]> list = new ArrayList<String[]>();
        String[] array = { "a", "b", "c", "ab", "cd", "abc", "abcde", "fghij", "klmno" };
        int srcPos;
        for (int i = 0; i < array.length; i++) {
            srcPos = i;
            while (i < array.length - 1 && array[i].length() == array[i + 1].length()) {
                i++;
            }
            String[] subarray = new String[i - srcPos + 1];
            System.arraycopy(array, srcPos, subarray, 0, subarray.length);
            list.add(subarray);
        }

        // Display the subarrays
        for (String[] arr : list) {
            System.out.println(Arrays.toString(arr));
        }
    }
}

输出：

[a, b, c]
[ab, cd]
[abc]
[abcde, fghij, klmno]

Answer 2

您可以使用Map 将字符串长度关联到该长度的字符串子数组：

String[] array = {"a", "b", "c", "ab", "cd", "abc", "abcde", "fghij", "klmno"};

Map<Integer, String[]> map = new HashMap<>();

for(int j=0, i=1; i<=array.length; i++)
{
    if(i == array.length || array[i].length() > array[j].length())
    {
        map.put(array[j].length(), Arrays.copyOfRange(array, j, i)) ;
        j = i;
    }
}

for(Integer len: map.keySet())
    System.out.format("%d : %s%n", len, Arrays.toString(map.get(len)));

输出：

1 : [a, b, c]
2 : [ab, cd]
3 : [abc]
5 : [abcde, fghij, klmno]

Answer 3

我的解决方案与@QuickSilver 的相同，只是不太清楚。 既然我在这里，我也放我的，因为我有专门的时间，但我再说一遍，我建议跟随他。

代码

public static void main(String[] args) {
        String[] array = {"a", "b", "c", "ab", "cd", "abc", "abcde", "fghij", "klmdfwetdfgdfgdfgdg"};
        HashMap<Integer, List<String>> hashMap = new HashMap<>();
        int strLength = array[0].length();

        for (String s : array) {
            while (true) {
                if (s.length() == strLength) {
                    if (hashMap.get(strLength) != null) {
                        List<String> temp = hashMap.get(strLength);
                        temp.add(s);
                        hashMap.put(strLength, temp);
                    } else {
                        List<String> strings = new LinkedList<>();
                        strings.add(s);
                        hashMap.put(strLength, strings);
                    }
                    break;
                } else
                    strLength = s.length();
            }
        }
        System.out.println(hashMap);
    }

Answer 4

为了快速访问，您还可以使用列表列表。

String[] array = new String[]{"a","b","c","ab","cd","abc","abcde","fghij","klmno"};
List<List<String>> lists = new LinkedList<>();

// you will have to update this number based on the maximum length of string you are expecting
for (int i = 0; i < 6; i++) {
    lists.add(new LinkedList<>());
}

for (String a: array) {
    lists.get(a.length()).add(a);
}

System.out.println(lists);

这里，第一个列表用于大小，内部列表用于实际字符串。

注意：这仅适用于较小的字符串。如果您有长度为 1、2、100 的字符串。您可能应该使用 HashMaps，因为这种方法会浪费大量内存。

---

使用 Java8：

String[] array = new String[]{"a","b","c","ab","cd","abc","abcde","fghij","klmno"};

List<List<String>> lists = IntStream.range(0, 6).<List<String>>mapToObj(
    i -> new LinkedList<>()).collect(Collectors.toCollection(LinkedList::new));

Arrays.stream(array).forEach(a -> lists.get(a.length()).add(a));

System.out.println(lists);

Answer 5

最好创建一个Map<Integer, List<String>>，其中键是字符串的长度，值是类似大小的字符串列表。

import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.Map;

public class SimpleArray {

    public static void main(String[] args) {
        String[] array = new String[]{"a","b","c","ab","cd","abc","abcde","fghij","klmno"};

        Map<Integer, List<String>> map = new HashMap<>();

        for (int i = 0; i < array.length; i++) {
            List< String> temp = map.getOrDefault(array[i].length(),new ArrayList<>());
            temp.add(array[i]);
            map.put(array[i].length(),temp);
        }
        System.out.println(map);

    }
}