[leetcode]Shortest Distance from All Buildings
首先用flooding,暴力的多次bfs,差一点就要超时
from queue import Queue class Solution: def markDistance(self, grid, i, j): m = len(grid) n = len(grid[0]) distance = [[None] * n for k in range(m)] que = Queue() distance[i][j] = 0 que.put((i, j)) while not que.empty(): x, y = que.get() current = distance[x][y] if grid[x][y] == 1 and current != 0: continue for dx, dy in [(0, 1), (0, -1), (1, 0), (-1, 0)]: if x + dx < 0 or x + dx >= m or y + dy < 0 or y + dy >= n: continue if grid[x + dx][y + dy] == 2: # block continue if distance[x + dx][y + dy] is not None: # already visited continue distance[x + dx][y + dy] = current + 1 que.put((x + dx, y + dy)) return distance def shortestDistance(self, grid: List[List[int]]) -> int: if not len(grid) or not len(grid[0]): return -1 m = len(grid) n = len(grid[0]) buildings = [] for i in range(m): for j in range(n): if grid[i][j] == 1: buildings.append((i, j)) distances = [] # None for not visited, int for distances for building in buildings: distances.append(self.markDistance(grid, building[0], building[1])) result = -1 for i in range(m): for j in range(n): current = 0 for k in range(len(distances)): if not distances[k][i][j]: current = -1 break else: current += distances[k][i][j] if current != -1 and (result == -1 or result > current): result = current return result