力扣317python

力扣317题的Python解法如下：

from collections import deque

def shortestDistance(grid):
    if not grid or not grid[0]:
        return -1

    rows, cols = len(grid), len(grid[0])
    buildings = 0
    distance = [[0] * cols for _ in range(rows)]
    reach = [[0] * cols for _ in range(rows)]

    for i in range(rows):
        for j in range(cols):
            if grid[i][j] == 1:
                buildings += 1
                if not bfs(grid, distance, reach, i, j):
                    return -1

    min_distance = float('inf')
    for i in range(rows):
        for j in range(cols):
            if grid[i][j] == 0 and reach[i][j] == buildings:
                min_distance = min(min_distance, distance[i][j])

    return min_distance if min_distance != float('inf') else -1

def bfs(grid, distance, reach, row, col):
    rows, cols = len(grid), len(grid[0])
    visited = [[False] * cols for _ in range(rows)]
    queue = deque([(row, col)])
    visited[row][col] = True

    level = 0
    while queue:
        level += 1
        size = len(queue)

        for _ in range(size):
            x, y = queue.popleft()
            neighbors = [(x-1, y), (x+1, y), (x, y-1), (x, y+1)]

            for nx, ny in neighbors:
                if 0 <= nx < rows and 0 <= ny < cols and not visited[nx][ny]:
                    visited[nx][ny] = True
                    if grid[nx][ny] == 0:
                        distance[nx][ny] += level
                        reach[nx][ny] += 1
                        queue.append((nx, ny))
                    elif grid[nx][ny] == 1:
                        continue
                    else:
                        return False  # obstacle

    return True

这段代码用于解决力扣317题，是一个基于广度优先搜索的解法。它计算了每个空地到所有建筑物的最短距离，并找到距离所有建筑物最近的空地。如果有多个最近的空地，则返回其中任意一个的距离。如果不存在这样的空地，则返回-1。