leetcode1030

class Solution(object):
    def __init__(self):
        self.List = list()

    def bfs(self,R,C,S,V):
        T = list()
        while len(S) >0:
            node = S.pop(0)
            if node[0]-1>=0 and V[node[0]-1][node[1]] == 0:
                V[node[0]-1][node[1]] = 1
                T.append([node[0]-1,node[1]])
                self.List.append([node[0]-1,node[1]])
            if node[0]+1<R and V[node[0]+1][node[1]] == 0:
                V[node[0]+1][node[1]] = 1
                T.append([node[0]+1,node[1]])
                self.List.append([node[0]+1,node[1]])
            if node[1]-1>=0 and V[node[0]][node[1]-1] == 0:
                V[node[0]][node[1]-1] = 1
                T.append([node[0],node[1]-1])
                self.List.append([node[0],node[1]-1])
            if node[1]+1<C and V[node[0]][node[1]+1] == 0:
                V[node[0]][node[1]+1] = 1
                T.append([node[0],node[1]+1])
                self.List.append([node[0],node[1]+1])
        if len(T)>0:
            self.bfs(R,C,T,V)



    def allCellsDistOrder(self, R: int, C: int, r0: int, c0: int) -> 'List[List[int]]':
        stack = list()
        visited = [[0 for col in range(C)] for row in range(R)]
        stack.append([r0,c0])
        visited[r0][c0] = 1
        self.List.append([r0,c0])
        self.bfs(R,C,stack,visited)
        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。