Spiral Matrix LeetCode Solution | Easy Approach

Minimum Cost to Merge Stones
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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 Solutions

Time: O(mn)
Space: O(1)

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;
  }
};

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;
  }
}

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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