Print Matrix Diagonally - Google Interview

http://algorithms.tutorialhorizon.com/print-all-diagonals-of-a-given-matrix/

We will solve this prob­lem in two parts. first half of diag­o­nals and sec­ond half of diagonals.

public static void print(int [][] a){

    //print first half

    int row =0;

    int col;

    while(row<a.length){

        col =0;

        int rowTemp = row;

        while(rowTemp>=0){

            System.out.print(a[rowTemp][col] + " ");

            rowTemp--;

            col++;

        }

        System.out.println();

        row++;

    }

    //print second half

    col = 1;

    while(col<a.length){

        int colTemp = col;

        row = a.length-1;

        while(colTemp<=a.length-1){

            System.out.print(a[row][colTemp] + " ");

            row--;

            colTemp++;

        }

        System.out.println();

        col++;

    }

}

Google Interview - Print Matrix Diagonally - 我的博客 - ITeye技术网站
Given a 2D matrix, print all elements of the given matrix in diagonal order.

    1     2     3     4
    5     6     7     8
    9    10    11    12
   13    14    15    16
   17    18    19    20

Diagonal printing of the above matrix is

    1
    5     2
    9     6     3
   13    10     7     4
   17    14    11     8
   18    15    12
   19    16
   20

1. public void printDiagonal(int[][] A) {  
2.     int m = A.length, n = A[0].length;  
3.     for (int i = 0; i < m+n-1; i++) {  
4.         int row = Math.min(i, m-1);  
5.         int col = Math.max(0, i-m+1);  
6.         while(row>=0 && col<n) {  
7.             System.out.print(A[row--][col++] + " ");  
8.         }  
9.         System.out.println();  
10.     }  
11. } 

http://segmentfault.com/a/1190000003890376

总共需要打印的层数，是长度加宽度减去一。关键在于内层的row = i - j，而col = j。

private static void printDiagonal(int[][] matrix){
    int m = matrix.length;
    int n = matrix[0].length;
    // 计算打印的层数
    int lvl = m + n - 1;
    for(int i = 0; i < lvl; i++){
        for(int j = i; j >= 0; j--){
            int row = i - j;
            int col = j;
            // 超过边界的点直接跳过
            if(row >= m || col >= n) continue;
            System.out.print(matrix[row][col]+" ");
        }
        System.out.println();
    }
}

http://www.ideserve.co.in/learn/print-matrix-diagonally
Given a matrix of mxn dimensions, print the elements of the matrix in diagonal order.

Then, number of diagonals will be = rowCount + columnCount - 1 as depicted in the diagram below.

    public static void printMatrixDiagonally(int[][] matrix) {

        int row, col;

        int rowCount = matrix.length;

        int columnCount = matrix[0].length;

        for (int k = 0; k < rowCount; k++) {

            for (row = k, col = 0; row >= 0 && col < columnCount; row--, col++) {

                System.out.print(matrix[row][col] + " ");

            }

            System.out.println();

        }

        for (int k = 1; k < columnCount; k++) {

            for (row = rowCount - 1, col = k; row >= 0 && col < columnCount; row--, col++) {

                System.out.print(matrix[row][col] + " ");

            }

            System.out.println();

        }

    }

http://www.geeksforgeeks.org/print-matrix-diagonal-pattern/

Given a matrix of n*n size, the task is to print its elements in diagonally pattern.

1. We use a flag isUp to decide whether we need to go upward direction or downward direction. We set isUp = true initially that first we are going upward.
2. If isUp = 1 then start printing elements by incrementing column index and decrementing the row index.
3. Similarly if isUp = 0, then decrement the column index and increment the row index.
5. Do this till all the elements get traversed.

void printMatrixDiagonal(int mat[MAX][MAX], int n)

{

    // Initialize indexes of element to be printed next

    int i = 0, j = 0 ;

    // Direction is initially from down to up

    bool isUp = true;

    // Traverse the matrix till all elements get traversed

    for (int k=0; k<n*n;)

    {

        // If isUp = true then traverse from downward

        // to upward

        if (isUp)

        {

            for ( ; i>=0 && j<n ; j++, i--)

            {

                cout << mat[i][j] << " ";

                k++;

            }

            // Set i and j according to direction

            if (i < 0 && j<=n-1)

                i = 0;

            if (j == n)

                i = i+2 , j--;

        }

        // If isUp = 0 then traverse up to down

        else

        {

            for ( ; j >= 0 && i<n ; i++ , j--)

            {

                cout << mat[i][j] << " ";

                k++;

            }

            // Set i and j according to direction

            if (j < 0 && i<=n-1)

                j = 0;

            if (i == n)

                j = j + 2 , i--;

        }

        // Revert the isUp to change the direction

        isUp = !isUp;

    }

}

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