Spiral Matrix LeetCode Programming Solutions | LeetCode Problem Solutions in C++, Java, & Python [💯Correct]

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In this post, you will find the solution for the Spiral Matrix in C++, Java & Python-LeetCode problem. We are providing the correct and tested solutions to coding problems present on LeetCode. If you are not able to solve any problem, then you can take help from our Blog/website.

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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:

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 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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