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# Spiral Matrix LeetCode Solution

## Problem – Spiral Matrix

Given an `m x n` `matrix`, return all elements of the `matrix` in spiral order.

Example 1:

``````Input: matrix = [[1,2,3],[4,5,6],[7,8,9]]
Output: [1,2,3,6,9,8,7,4,5]``````

Example 2:

``````Input: matrix = [[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]``````

Constraints:

• `m == matrix.length`
• `n == matrix[i].length`
• `1 <= m, n <= 10`
• `-100 <= matrix[i][j] <= 100`

### Spiral Matrix LeetCode Solution in Java

``````public class Solution {
public List<Integer> spiralOrder(int[][] matrix) {

List<Integer> res = new ArrayList<Integer>();

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

int rowBegin = 0;
int rowEnd = matrix.length-1;
int colBegin = 0;
int colEnd = matrix[0].length - 1;

while (rowBegin <= rowEnd && colBegin <= colEnd) {
// Traverse Right
for (int j = colBegin; j <= colEnd; j ++) {
}
rowBegin++;

// Traverse Down
for (int j = rowBegin; j <= rowEnd; j ++) {
}
colEnd--;

if (rowBegin <= rowEnd) {
// Traverse Left
for (int j = colEnd; j >= colBegin; j --) {
}
}
rowEnd--;

if (colBegin <= colEnd) {
// Traver Up
for (int j = rowEnd; j >= rowBegin; j --) {
}
}
colBegin ++;
}

return res;
}
}
``````

### Spiral Matrix LeetCode Solution in Python

``````def spiralOrder(self, matrix):
return matrix and list(matrix.pop(0)) + self.spiralOrder(zip(*matrix)[::-1])``````

### Spiral Matrix LeetCode Solution in C++

``````vector<int> spiralOrder(vector<vector<int>>& matrix) {
vector<vector<int> > dirs{{0, 1}, {1, 0}, {0, -1}, {-1, 0}};
vector<int> res;
int nr = matrix.size();     if (nr == 0) return res;
int nc = matrix[0].size();  if (nc == 0) return res;

vector<int> nSteps{nc, nr-1};

int iDir = 0;   // index of direction.
int ir = 0, ic = -1;    // initial position
while (nSteps[iDir%2]) {
for (int i = 0; i < nSteps[iDir%2]; ++i) {
ir += dirs[iDir][0]; ic += dirs[iDir][1];
res.push_back(matrix[ir][ic]);
}
nSteps[iDir%2]--;
iDir = (iDir + 1) % 4;
}
return res;
}
``````
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