Leetcode 1030. Matrix Cells in Distance Order

文章作者：Tyan 博客：noahsnail.com | CSDN | 简书

1. Description

2. Solution

**解析：**Version 1，直接根据距离进行排序。Version 使用广度优先搜索。

• Version 1

代码语言：javascript复制

class Solution:
    def allCellsDistOrder(self, rows: int, cols: int, rCenter: int, cCenter: int) -> List[List[int]]:
        coordinates = [[i, j] for i in range(rows) for j in range(cols)]
        coordinates.sort(key=lambda x: abs(x[0] - rCenter)   abs(x[1] - cCenter))
        return coordinates

• Version 2

代码语言：javascript复制

class Solution:
    def allCellsDistOrder(self, rows: int, cols: int, rCenter: int, cCenter: int) -> List[List[int]]:
        matrix = [[0] * cols for i in range(rows)]
        queue = collections.deque()
        queue.append((rCenter, cCenter))
        matrix[rCenter][cCenter] = 1
        coordinates = []
        while queue:
            x, y = queue.popleft()
            coordinates.append([x, y])
            matrix[x][y] = 1
            if x > 0 and matrix[x-1][y] == 0:
                matrix[x-1][y] = 1
                queue.append((x-1, y))             
            if x < rows - 1 and matrix[x 1][y] == 0:
                matrix[x 1][y] = 1
                queue.append((x 1, y))   
            if y > 0 and matrix[x][y-1] == 0:
                matrix[x][y-1] = 1
                queue.append((x, y-1))             
            if y < cols - 1 and matrix[x][y 1] == 0:
                matrix[x][y 1] = 1
                queue.append((x, y 1))   
        return coordinates

Reference

1. https://leetcode.com/problems/matrix-cells-in-distance-order/

sdn com version 博客 排序

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