# 54. Spiral Matrix (Medium)

https://leetcode.com/problems/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]
```

## Solutions

``````class Solution {
public List<Integer> spiralOrder(int[][] matrix) {
List<Integer> ans =new ArrayList<>();

if (matrix.length == 0) {
return ans;
}

int row = matrix.length;
int col = matrix[0].length;

int m = Math.min(col, row);

for (int i = 0; i < (int) Math.ceil(m / 2.0); i++) {
List<Integer> tmp = spiralOrder(matrix, i);
}

return ans;
}

private List<Integer> spiralOrder(int[][] matrix, int margin) {
List<Integer> ans =new ArrayList<>();

int row = matrix.length;
int col = matrix[0].length;

// Literally, following case will never matched, just for reuse purpose
if (margin >= col || margin >= row) {
return ans;
}

// Go through top row
for (int i = margin; i < col - margin; i++) {
}

// Go through right column
for (int i = margin + 1; i < row - margin; i++) {
}

// Go through bottom row, in case the spiral only contains one row
for (int i = col - margin - 2; i >= margin && (row - margin - 1 > margin); i--) {
}

// Go through left column, in case the spiral only contains one column
for (int i = row - margin - 2; i > margin && (col - margin - 1 > margin); i--) {
}

return ans;
}
}
``````