package y2019.Algorithm.array;

/**

 * @ProjectName: cutter-point

 * @Package: y2019.Algorithm.array

 * @ClassName: MatrixReshape

 * @Author: xiaof

 * @Description:  TODO 566. Reshape the Matrix

 *

 * In MATLAB, there is a very useful function called 'reshape',

 * which can reshape a matrix into a new one with different size but keep its original data.

 * You're given a matrix represented by a two-dimensional array,

 * and two positive integers r and c representing the row number and column number of the wanted reshaped matrix, respectively.

 * The reshaped matrix need to be filled with all the elements of the original matrix in the same row-traversing order as they were.

 * If the 'reshape' operation with given parameters is possible and legal, output the new reshaped matrix; Otherwise,

 * output the original matrix.

 *

 * Input:

 * nums =

 * [[1,2],

 *  [3,4]]

 * r = 1, c = 4

 * Output:

 * [[1,2,3,4]]

 * Explanation:

 * The row-traversing of nums is [1,2,3,4]. The new reshaped matrix is a 1 * 4 matrix, fill it row by row by using the previous list.

 *

 * @Date: 2019/7/8 9:03

 * @Version: 1.0

 */

public class MatrixReshape {

    public int[][] solution(int[][] nums, int r, int c) {

        //吧对应的数组输出为新的矩阵数组，如果不够那么直接输出原来数组

        if(nums[0].length * nums.length != (r * c)) {

            return nums;

        }

        int[][] result = new int[r][c];

        int indexNum = 0;

        //依次遍历数组

        for(int i = 0; i < nums.length; ++i) {

            for(int j = 0; j < nums[i].length; ++j) {

                //遍历所有

                int curNum = i * nums[i].length + j;

                result[curNum / c][curNum % c] = nums[i][j];

            }

        }

        return result;

    }

    public static void main(String args[]) {

        int A[][] = {{1,2},{3,4}};

        MatrixReshape fuc = new MatrixReshape();

        System.out.println(fuc.solution(A, 1, 4));

    }

}

package y2019.Algorithm.array;

import java.util.Arrays;

/**

 * @ProjectName: cutter-point

 * @Package: y2019.Algorithm.array

 * @ClassName: DuplicateZeros

 * @Author: xiaof

 * @Description: TODO 1089. Duplicate Zeros

 * Given a fixed length array arr of integers, duplicate each occurrence of zero, shifting the remaining elements to the right.

 * Note that elements beyond the length of the original array are not written.

 * Do the above modifications to the input array in place, do not return anything from your function.

 *

 * Input: [1,0,2,3,0,4,5,0]

 * Output: null

 * Explanation: After calling your function, the input array is modified to: [1,0,0,2,3,0,0,4]

 *

 * @Date: 2019/7/8 9:17

 * @Version: 1.0

 */

public class DuplicateZeros {

    public void solution(int[] arr) {

        //每次遇到0就修改为两次0，然后所有其他的数据右移

        int[] copyArry = Arrays.copyOf(arr, arr.length);

        int index = 0;

        for(int i = 0; i < copyArry.length && index < arr.length; ++i) {

            if(copyArry[i] == 0) {

                arr[index++] = 0;

                if(index < arr.length)

                    arr[index++] = 0;

            } else {

                arr[index++] = copyArry[i];

            }

        }

    }

}