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 ++) {
                res.add(matrix[rowBegin][j]);
            }
            rowBegin++;
            
            // Traverse Down
            for (int j = rowBegin; j <= rowEnd; j ++) {
                res.add(matrix[j][colEnd]);
            }
            colEnd--;
            
            if (rowBegin <= rowEnd) {
                // Traverse Left
                for (int j = colEnd; j >= colBegin; j --) {
                    res.add(matrix[rowEnd][j]);
                }
            }
            rowEnd--;
            
            if (colBegin <= colEnd) {
                // Traver Up
                for (int j = rowEnd; j >= rowBegin; j --) {
                    res.add(matrix[j][colBegin]);
                }
            }
            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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