Question

如何从索引位置移动二维数组

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任何想法？感谢

Answer 1

这是使用Guava和Java 8的一种方法：

for (int[] array : x) {
    Collections.rotate(Ints.asList(array), index);
}

或Java 7版本：

ID: COM-1234
Program: Swimming
Name: John Doe
Description: Joined on July 1st
------------------------------ID: COM-2345
Program: Swimming
Name: Brock sen
Description: Joined on July 1st
------------------------------ID: COM-9876
Program: Swimming
Name: johny boy
Description: Joined on July 1st
------------------------------ID: COM-9090
Program: Running
Name: justin kim
Description: Good Record
------------------------------

Answer 2

这个程序使用一个逻辑，我们通过在部分中反转数组来旋转数组。首先，我们将数组反转到索引位置，然后反转剩余的数组（索引+ 1到数组的最后一个元素）。在完成上述两个步骤之后，我们再次调用反向函数，但这次是在整个数组上，它为我们提供了所需的输出。

以下是有助于理解上述逻辑的代码。

public class ShiftTwoDArray {

    public static void main(String[] args) {
        int[][] x = {   { 1, 2, 3, 4, 5, 6, 7 }, 
                        { 1, 2, 3, 4, 5, 6, 7 }, 
                        { 1, 2, 3, 4, 5, 6, 7 },
                        { 1, 2, 3, 4, 5, 6, 7 } 
                    };

        int index = 3;
        int i, j;
        // System.out.println(x.length);
        System.out.println("Before");
        for (i = 0; i < x.length; i++) {
            for (j = 0; j < x[i].length; j++) {
                System.out.print(x[i][j] + "  ");
            }
            System.out.println();
        }
        rotate(x, index);

        System.out.println("\nAfter");
        for (i = 0; i < x.length; i++) {
            for (j = 0; j < x[i].length; j++) {
                System.out.print(x[i][j] + "  ");
            }
            System.out.println();
        }
    }

    /**
     * @param x
     * @param index
     * calls rotateUtil on each row
     */
    private static void rotate(int[][] x, int index) {
        for (int i = 0; i < x.length; i++) {
            rotateUtil(x[i], index);
        }
    }

    /**
     * @param x
     * @param index
     * reverse array in parts and then reverse whole array
     */
    private static void rotateUtil(int[] x, int index) {
        reverse(x, 0, index);
        reverse(x, index + 1, x.length - 1);
        reverse(x, 0, x.length - 1);
    }

    /**
     * @param x
     * @param start
     * @param end
     * reverse an array
     */
    private static void reverse(int[] x, int start, int end) {
        int temp = 0;
        while (start < end) {
            temp = x[start];
            x[start] = x[end];
            x[end] = temp;
            start++;
            end--;
        }
    }
}

Answer 3

这应该做你想要的：

int[][] x = { { 1, 2, 3, 4, 5, 6, 7 }, 
              { 1, 2, 3, 4, 5, 6, 7 },
              { 1, 2, 3, 4, 5, 6, 7 },
              { 1, 2, 3, 4, 5, 6, 7 } };

int index = 3;

// Create 2D array of matching size
int[][] y = new int[x.length][x[0].length];

// Loop through each row of x
for (int r = 0; r < x.length; r++) {
    // Loop through each column of x[r][]
    for (int c = 0; c < x[0].length; c++) {
        // Put x's value in y, shifting to the right by index.
        // See comment outside of code regarding %
        y[r][(c + index) % x[0].length] = x[r][c];
    }
}

// Print out y to see if it worked
for (int r = 0; r < y.length; r++) {
    for (int c = 0; c < y[0].length; c++) {
        System.out.print(y[r][c] + " ");
    }
    System.out.println();
}

此处的关键是% x[0].length中的y[r][(c + index) % x[0].length] = x[r][c];。 c + index将列向右移动。但是，如果需要，% x[0].length会将列包裹到行的开头。

按列索引移动二维数组中的所有元素

3 个答案: