feat(54-spiral-matrix): add py3 solution
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0054_spiral-matrix/README.md
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0054_spiral-matrix/README.md
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Given an `m x n` `matrix`, return _all elements of the_ `matrix` _in spiral order_.
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Example 1:
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![](https://assets.leetcode.com/uploads/2020/11/13/spiral1.jpg)
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Input: matrix = [[1,2,3],[4,5,6],[7,8,9]] Output: [1,2,3,6,9,8,7,4,5]
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Example 2:
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![](https://assets.leetcode.com/uploads/2020/11/13/spiral.jpg)
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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]
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Constraints:
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* `m == matrix.length`
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* `n == matrix[i].length`
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* `1 <= m, n <= 10`
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* `-100 <= matrix[i][j] <= 100`
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0054_spiral-matrix/python3/solution.py
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0054_spiral-matrix/python3/solution.py
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class Solution:
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def spiralOrder(self, matrix: List[List[int]]) -> List[int]:
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row_begin_loc, row_end_loc = 0, len(matrix) - 1
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col_begin_loc, col_end_loc = 0, len(matrix[0]) - 1
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result = []
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while row_begin_loc <= row_end_loc and col_begin_loc <= col_end_loc:
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# Top Side
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i, j = row_begin_loc, col_begin_loc
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while j <= col_end_loc:
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result.append(matrix[i][j])
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j += 1
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# Right Side
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i += 1
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j -= 1
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while i <= row_end_loc:
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result.append(matrix[i][j])
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i += 1
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# Bottom Side
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i -= 1
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j -= 1
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while j >= col_begin_loc and i > row_begin_loc:
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result.append(matrix[i][j])
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j -= 1
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# Left Side
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i -= 1
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j += 1
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while i > row_begin_loc and j < col_end_loc:
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result.append(matrix[i][j])
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i -= 1
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row_begin_loc += 1
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row_end_loc -= 1
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col_begin_loc += 1
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col_end_loc -= 1
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return result
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