add spiralOrder function to return elements of a matrix in spiral order

Author: jamiebase

spiral-matrix/gcount85.py

@@ -0,0 +1,41 @@
+"""
+# Approach
+상하좌우 경계값을 설정하고 matrix을 순회하며 경계값을 줄여나갑니다.
+
+# Complexity
+matrix 크기의 가로를 M, 세로를 N이라고 할 때
+- Time complexity: O(M*N)
+- Space complexity: O(M*N)
+"""
+
+
+class Solution:
+    def spiralOrder(self, matrix: list[list[int]]) -> list[int]:
+        start_row, start_col = 0, 0
+        end_row, end_col = len(matrix) - 1, len(matrix[0]) - 1
+        output = []
+
+        while start_row <= end_row and start_col <= end_col:
+            # 좌->우
+            for col in range(start_col, end_col + 1):
+                output.append(matrix[start_row][col])
+            start_row += 1
+
+            # 상->하
+            for row in range(start_row, end_row + 1):
+                output.append(matrix[row][end_col])
+            end_col -= 1
+
+            # 우->좌
+            if start_row <= end_row:
+                for col in range(end_col, start_col - 1, -1):
+                    output.append(matrix[end_row][col])
+                end_row -= 1
+
+            # 하->상
+            if start_col <= end_col:
+                for row in range(end_row, start_row - 1, -1):
+                    output.append(matrix[row][start_col])
+                start_col += 1
+
+        return output