Spiral Matrix

Given a matrix of m x n elements (m rows, n columns), return all elements of the matrix in spiral order.

For example,
Given the following matrix:

  1. [
  2.  [ 1, 2, 3 ],
  3.  [ 4, 5, 6 ],
  4.  [ 7, 8, 9 ]
  5. ]

You should return [1,2,3,6,9,8,7,4,5].

Solution:

  1. public class Solution {
  2.   public List<Integer> spiralOrder(int[][] matrix) {
  3.     List<Integer> res = new ArrayList<Integer>();
  5.     if (matrix == null || matrix.length == 0 || matrix[0].length == 0) {
  6.       return res;
  7.     }
  9.     int m = matrix.length;
  10.     int n = matrix[0].length;
  12.     int layers = (int)Math.ceil(Math.min(m, n) / 2.0);
  14.     for (int k = 0; k < layers; k++) {
  15.       if (m <= 0 || n <= 0) {
  16.         break;
  17.       }
  19.       // one row
  20.       if (m == 1) {
  21.         for (int j = 0; j < n; j++) {
  22.           res.add(matrix[k][k + j]);
  23.         }
  24.       }
  26.       // one column
  27.       else if (n == 1) {
  28.         for (int i = 0; i < m; i++) {
  29.           res.add(matrix[k + i][k]);
  30.         }
  31.       }
  33.       else {
  34.         // top
  35.         for (int j = 0; j < n - 1; j++) {
  36.           res.add(matrix[k][k + j]);
  37.         }
  39.         // right
  40.         for (int i = 0; i < m - 1; i++) {
  41.           res.add(matrix[k + i][k + n - 1]);
  42.         }
  44.         // bottom
  45.         for (int j = n - 1; j > 0; j--) {
  46.           res.add(matrix[k + m - 1][k + j]);
  47.         }
  49.         // left
  50.         for (int i = m - 1; i > 0; i--) {
  51.           res.add(matrix[k + i][k]);
  52.         }
  53.       }
  55.       m -= 2;
  56.       n -= 2;
  57.     }
  59.     return res;
  60.   }
  61. }