In this post, you will know how to solve the Spiral Matrix Leetcode Solution problem of Leetcode. This Leetcode problem is done in many programming languages like C++, Java, and Python.Spiral Matrix Leetcode SolutionsOne more thing to add, don’t directly look for the solutions, first try to solve the problems of Leetcode by yourself. If you find any difficulty after trying several times, then you can look for solutions.ProblemGiven 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.lengthn == matrix[i].length1 <= m, n <= 10-100 <= matrix[i][j] <= 100Spiral Matrix Leetcode Solutions in Pythonclass Solution: def spiralOrder(self, matrix: List[List[int]]) -> List[int]: if not matrix: return [] m = len(matrix) n = len(matrix[0]) ans = [] r1 = 0 c1 = 0 r2 = m - 1 c2 = n - 1 # Repeatedly add matrix[r1..r2][c1..c2] to ans while len(ans) < m * n: j = c1 while j <= c2 and len(ans) < m * n: ans.append(matrix[r1][j]) j += 1 i = r1 + 1 while i <= r2 - 1 and len(ans) < m * n: ans.append(matrix[i][c2]) i += 1 j = c2 while j >= c1 and len(ans) < m * n: ans.append(matrix[r2][j]) j -= 1 i = r2 - 1 while i >= r1 + 1 and len(ans) < m * n: ans.append(matrix[i][c1]) i -= 1 r1 += 1 c1 += 1 r2 -= 1 c2 -= 1 return ans Spiral Matrix Leetcode Solutions in CPPclass Solution { public: vector<int> spiralOrder(vector<vector<int>>& matrix) { if (matrix.empty()) return {}; const int m = matrix.size(); const int n = matrix[0].size(); vector<int> ans; int r1 = 0; int c1 = 0; int r2 = m - 1; int c2 = n - 1; // Repeatedly add matrix[r1..r2][c1..c2] to ans while (ans.size() < m * n) { for (int j = c1; j <= c2 && ans.size() < m * n; ++j) ans.push_back(matrix[r1][j]); for (int i = r1 + 1; i <= r2 - 1 && ans.size() < m * n; ++i) ans.push_back(matrix[i][c2]); for (int j = c2; j >= c1 && ans.size() < m * n; --j) ans.push_back(matrix[r2][j]); for (int i = r2 - 1; i >= r1 + 1 && ans.size() < m * n; --i) ans.push_back(matrix[i][c1]); ++r1, ++c1, --r2, --c2; } return ans; } }; Spiral Matrix Leetcode Solutions in Javaclass Solution { public List<Integer> spiralOrder(int[][] matrix) { if (matrix.length == 0) return new ArrayList<>(); final int m = matrix.length; final int n = matrix[0].length; List<Integer> ans = new ArrayList<>(); int r1 = 0; int c1 = 0; int r2 = m - 1; int c2 = n - 1; // Repeatedly add matrix[r1..r2][c1..c2] to ans while (ans.size() < m * n) { for (int j = c1; j <= c2 && ans.size() < m * n; ++j) ans.add(matrix[r1][j]); for (int i = r1 + 1; i <= r2 - 1 && ans.size() < m * n; ++i) ans.add(matrix[i][c2]); for (int j = c2; j >= c1 && ans.size() < m * n; --j) ans.add(matrix[r2][j]); for (int i = r2 - 1; i >= r1 + 1 && ans.size() < m * n; --i) ans.add(matrix[i][c1]); ++r1; ++c1; --r2; --c2; } return ans; } } Note: This problem Spiral Matrix is generated by Leetcode but the solution is provided by BrokenProgrammers. This tutorial is only for Educational and Learning purposes.Next: Partition List Leetcode Solution Post navigationScramble String Leetcode Solution Partition List Leetcode Solution