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Link for the Problem – Spiral Matrix– LeetCode Problem
Spiral Matrix– LeetCode Problem
Problem:
Given an m x n
matrix
, return all elements of the matrix
in spiral order.
Example 1:
![Spiral Matrix LeetCode Programming Solutions | LeetCode Problem Solutions in C++, Java, & Python [💯Correct] 2 spiral1](https://assets.leetcode.com/uploads/2020/11/13/spiral1.jpg)
Input: matrix = [[1,2,3],[4,5,6],[7,8,9]] Output: [1,2,3,6,9,8,7,4,5]
Example 2:
![Spiral Matrix LeetCode Programming Solutions | LeetCode Problem Solutions in C++, Java, & Python [💯Correct] 3 spiral](https://assets.leetcode.com/uploads/2020/11/13/spiral.jpg)
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 Solutions
Spiral Matrix in C++:
class 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 in Java:
class 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; } }
Spiral Matrix in Python:
class 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
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