Question

我遇到的问题是我有一个数组

String[] purposeAll= {"P1","P2"};

和数组二维

String[][] purpose = { 
        { "P1", "a", "b", "c", "d", "e", "f", "h" },
        { "P2", "a", "b", "c", "d", "e", "g", "i" },
        { "P3", "l", "m", "n", "o", "p", "q" } };

我需要在数组二维中搜索一维数组的每个索引，结果将得到：

arrayOfPurpose[][]={ 
        { "P1", "a", "b", "c", "d", "e", "f", "h" },
        { "P2", "a", "b", "c", "d", "e", "g", "i" }  };

我的代码是

public static void ArrayOfPurpose(String[][] purpose, String[] purposeAll) {

String[][][] arrayOfPurpose = new String[purposeAll.length][purpose.length][];
    ArrayList<String[]> list = new ArrayList<String[]>();

    for (int i = 0; i < purpose.length; i++) {
        list.add(purpose[i]);
    }
    String[] tmp;
    for (int k = 0; k < purposeAll.length; k++) {
        arrayOfPurpose[k]= new String[purposeAll.length][];
        System.out.print("ok");
        for (int i = 0; i < list.size(); i++) {

            tmp = list.get(i);
            arrayOfPurpose[i]= new String[purposeAll.length][tmp.length];
            for (int j = 0; j < purpose[i].length; j++) {
                 if (tmp[0] == (purposeAll[k])) {
                    arrayOfPurpose[k][i][j] = tmp[j];
                  System.out.println(arrayOfPurpose[k][i][j]);
                }

            }

        }
    }
}

非常感谢你！：）

Answer 1

使用您在问题中的内容，以非Java8方式，您可以这样做：

for (String valueToCheck : purposeAll) {
    for (String[] values : purpose) {
        if (values[0].equals(valueToCheck)) {
            list.add(values);
        }
    }
}

String[][] arrayOfPurpose = new String[list.size()][];

int index = 0;

for (String[] a : list) {
    arrayOfPurpose[index++] = a;
}


System.out.println("Size of arrayOfPurpose: " + arrayOfPurpose.length);

Answer 2

Java8流解决方案：

public class blah {

    static String[] purposeAll= {"P1","P2"};

    static String[][] purpose = { 
            { "P1", "a", "b", "c", "d", "e", "f", "h" },
            { "P2", "a", "b", "c", "d", "e", "g", "i" },
            { "P3", "l", "m", "n", "o", "p", "q" } };
    public static String[][] arrayOfPurpose(String[][] purpose, String[] purposeAll) {
        return Arrays.asList(purpose).stream()
            .filter(arr -> Arrays.asList(purposeAll).contains(arr[0]))
            .toArray(String[][]::new);
    }
    public static void main(String[] args) {
        System.out.println(Arrays.deepToString(arrayOfPurpose(purpose, purposeAll)));
        // prints [[P1, a, b, c, d, e, f, h], [P2, a, b, c, d, e, g, i]]
    }
}

Answer 3

我希望我能正确理解这个问题。

    String[] purposeAll= {"P1","P2"};
    //String[] purposeAll= {"P3","P1"}; Testing Stuff.

    String[][] purpose = { 
    { "P1", "a", "b", "c", "d", "e", "f", "h" },
    { "P2", "a", "b", "c", "d", "e", "g", "i" },
    { "P3", "l", "m", "n", "o", "p", "q" } };

    List<String[]> arrayOfPurposes = new ArrayList<>();
    //Looping through the purposeAll to check purpose's first 
    //entry and seeing if it equals the first purposeAll value
    for(int i = 0; i < purpose.length; i++){
        for(int j = 0; j < purposeAll.length; j++){
            if(purpose[i][0].equals(purposeAll[j])){
                arrayOfPurposes.add(purpose[i]);
            }
        }
    }

    //One way to get the results into a String[][]
    String[][] result = {arrayOfPurposes.get(0),arrayOfPurposes.get(1)};

    //Displaying to console
    for(int i = 0; i < result.length; i++){
        for(int j = 0; j < result[i].length; j++){
            System.out.print(result[i][j]);
        }
        System.out.println();
    }

我觉得你应该使用List of String []而不是String [] []，因为结果的初始化非常简单。在大多数情况下，列表也更有效且更有帮助。

在二维数组中搜索一个Dimensional数组的索引并按行java返回

3 个答案: