Given a matrix of M x N elements (M rows, N columns), return all elements of the matrix in diagonal order as shown in the below image. Example: Input: [ [ 1, 2, 3 ], [ 4, 5, 6 ], [ 7, 8, 9 ] ] Output: [1,2,4,7,5,3,6,8,9] Explanation:
I don‘t think this is a hard problem. It is easy to figure out the walk pattern. Anyway...
Walk patterns:
- If out of
bottom border(row >= m) then row = m - 1; col += 2; change walk direction. - if out of
right border(col >= n) then col = n - 1; row += 2; change walk direction. - if out of
top border(row < 0) then row = 0; change walk direction. - if out of
left border(col < 0) then col = 0; change walk direction. - Otherwise, just go along with the current direction.
Time complexity: O(m * n), m = number of rows, n = number of columns.
Space complexity: O(1).
public class Solution {
public int[] findDiagonalOrder(int[][] matrix) {
if (matrix == null || matrix.length == 0) return new int[0];
int m = matrix.length, n = matrix[0].length;
int[] result = new int[m * n];
int row = 0, col = 0, d = 0;
int[][] dirs = {{-1, 1}, {1, -1}};
for (int i = 0; i < m * n; i++) {
result[i] = matrix[row][col];
row += dirs[d][0];
col += dirs[d][1];
if (row >= m) { row = m - 1; col += 2; d = 1 - d;}
if (col >= n) { col = n - 1; row += 2; d = 1 - d;}
if (row < 0) { row = 0; d = 1 - d;}
if (col < 0) { col = 0; d = 1 - d;}
}
return result;
}
}
