54. Spiral Matrix

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

Example 1:

Input:

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

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

Example 2:

Input:

[
  [1, 2, 3, 4],
  [5, 6, 7, 8],
  [9,10,11,12]
]

Output:
 [1,2,3,4,8,12,11,10,9,5,6,7]

Solution

(1) Java

(2) Python

class Solution:
    def spiralOrder(self, matrix):
        """
        :type matrix: List[List[int]]
        :rtype: List[int]
        """
        if not matrix or not matrix[0]:
            return []
        m = len(matrix)
        n = len(matrix[0])
        rst = []
        total = m*n
        u, d, l, r = 0, m, 0, n
        i, j = 0, 0
        while len(rst) < total:
            for j in range(l, r):
                rst.append(matrix[i][j])
            if len(rst) == total:
                break
            u += 1
            for i in range(u, d):
                rst.append(matrix[i][j])
            if len(rst) == total:
                break
            r -= 1
            for j in range(r-1, l-1, -1):
                rst.append(matrix[i][j])
            d -= 1
            for i in range(d-1, u-1, -1):
                rst.append(matrix[i][j])
            l += 1
        return rst

(3) Scala