1 class Solution(object): 2 def __init__(self): 3 self.List = list() 4 5 def bfs(self,R,C,S,V): 6 T = list() 7 while len(S) >0: 8 node = S.pop(0) 9 if node[0]-1>=0 and V[node[0]-1][node[1]] == 0: 10 V[node[0]-1][node[1]] = 1 11 T.append([node[0]-1,node[1]]) 12 self.List.append([node[0]-1,node[1]]) 13 if node[0]+1<R and V[node[0]+1][node[1]] == 0: 14 V[node[0]+1][node[1]] = 1 15 T.append([node[0]+1,node[1]]) 16 self.List.append([node[0]+1,node[1]]) 17 if node[1]-1>=0 and V[node[0]][node[1]-1] == 0: 18 V[node[0]][node[1]-1] = 1 19 T.append([node[0],node[1]-1]) 20 self.List.append([node[0],node[1]-1]) 21 if node[1]+1<C and V[node[0]][node[1]+1] == 0: 22 V[node[0]][node[1]+1] = 1 23 T.append([node[0],node[1]+1]) 24 self.List.append([node[0],node[1]+1]) 25 if len(T)>0: 26 self.bfs(R,C,T,V) 27 28 29 30 def allCellsDistOrder(self, R: int, C: int, r0: int, c0: int) -> 'List[List[int]]': 31 stack = list() 32 visited = [[0 for col in range(C)] for row in range(R)] 33 stack.append([r0,c0]) 34 visited[r0][c0] = 1 35 self.List.append([r0,c0]) 36 self.bfs(R,C,stack,visited) 37 return self.List
典型的BFS算法，每一“层”都比前一层的距离多1，因此按层遍历的顺序，即为所求。
