leetcode 1030. Matrix Cells in Distance Order ( Python )

描述

We are given a matrix with R rows and C columns has cells with integer coordinates (r, c), where 0 <= r < R and 0 <= c < C.

Additionally, we are given a cell in that matrix with coordinates (r0, c0).

Return the coordinates of all cells in the matrix, sorted by their distance from (r0, c0) from smallest distance to largest distance. Here, the distance between two cells (r1, c1) and (r2, c2) is the Manhattan distance, |r1 - r2| + |c1 - c2|. (You may return the answer in any order that satisfies this condition.)

Example 1:

Input: R = 1, C = 2, r0 = 0, c0 = 0
Output: [[0,0],[0,1]]
Explanation: The distances from (r0, c0) to other cells are: [0,1]
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Example 2:

Input: R = 2, C = 2, r0 = 0, c0 = 1
Output: [[0,1],[0,0],[1,1],[1,0]]
Explanation: The distances from (r0, c0) to other cells are: [0,1,1,2]
The answer [[0,1],[1,1],[0,0],[1,0]] would also be accepted as correct.
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Example 3:

Input: R = 2, C = 3, r0 = 1, c0 = 2
Output: [[1,2],[0,2],[1,1],[0,1],[1,0],[0,0]]
Explanation: The distances from (r0, c0) to other cells are: [0,1,1,2,2,3]
There are other answers that would also be accepted as correct, such as [[1,2],[1,1],[0,2],[1,0],[0,1],[0,0]].
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Note:

1 <= R <= 100
1 <= C <= 100
0 <= r0 < R
0 <= c0 < C
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解析

根據題意，就是計算矩陣中各個位置到 (r0, c0) 的曼哈頓距離，而後按照距離從小到大輸出矩陣的索引。時間複雜度爲 O(R * C)，空間複雜度爲 O(R * C)。

解答

class Solution(object):
    def allCellsDistOrder(self, R, C, r0, c0):
        """
        :type R: int
        :type C: int
        :type r0: int
        :type c0: int
        :rtype: List[List[int]]
        """
        def dist(point):
            pi, pj = point
            return abs(pi - r0) + abs(pj - c0)

        points = [(i, j) for i in range(R) for j in range(C)]
        return sorted(points, key=dist)				
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運行結果

Runtime: 132 ms, faster than 98.52% of Python online submissions for Matrix Cells in Distance Order.
Memory Usage: 13.6 MB, less than 99.10% of Python online submissions for Matrix Cells in Distance Order.
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