Leetcode: Walls and Gates
Question
You are given a m x n 2D grid initialized with these three possible values.
-1 - A wall or an obstacle.
0 - A gate.
INF - Infinity means an empty room. We use the value 231 - 1 = 2147483647 to represent INF as you may assume that the distance to a gate is less than 2147483647.
Fill each empty room with the distance to its nearest gate. If it is impossible to reach a gate, it should be filled with INF.
For example, given the 2D grid:
INF -1 0 INF
INF INF INF -1
INF -1 INF -1
0 -1 INF INF
After running your function, the 2D grid should be:
3 -1 0 1
2 2 1 -1
1 -1 2 -1
0 -1 3 4
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My first try
if there are k zero elements,
size(rooms) = (m,n)
Time complexity: O( kmn )
class Solution(object):
def wallsAndGates(self, rooms):
""" :type rooms: List[List[int]] :rtype: void Do not return anything, modify rooms in-place instead. """
if len(rooms)==0 or len(rooms[0])==0:
return
zero_list = []
for i in range(len(rooms)):
for j in range(len(rooms[0])):
if rooms[i][j]==0:
zero_list.append( (i,j) )
for zero_item in zero_list:
self.traverse4zero( zero_item, rooms )
def traverse4zero(self, zero_item, rooms):
#print "abc"
bfs_list = [zero_item]
dict_direct = [(-1,0), (1,0), (0,1), (0,-1)]
visited = [ [0]*len(rooms[0]) for dummy in range(len(rooms)) ]
visited[zero_item[0]][zero_item[1]] = 1
while bfs_list!=[]:
#print bfs_list
cur_item = bfs_list[0]
bfs_list = bfs_list[1:]
self.eachdirection( rooms, cur_item, visited, bfs_list, dict_direct )
return
def eachdirection(self, rooms, cur_item, visited, bfs_list, dict_direct):
for i in range(4):
next_i = cur_item[0] + dict_direct[i][0]
next_j = cur_item[1] + dict_direct[i][1]
if next_i<0 or next_i>=len(rooms) \
or next_j<0 or next_j>=len(rooms[0]) \
or rooms[next_i][next_j]==0 \
or rooms[next_i][next_j]==-1 \
or visited[next_i][next_j]==1 \
or rooms[cur_item[0]][cur_item[1]]+1 >= rooms[next_i][next_j]:
continue
rooms[next_i][next_j] = min( rooms[next_i][next_j], \
rooms[cur_item[0]][cur_item[1]]+1 )
bfs_list.append( (next_i, next_j) )
visited[next_i][next_j] = 1
return Solution
Accept
Time: 540ms
class Solution(object):
def wallsAndGates(self, rooms):
""" :type rooms: List[List[int]] :rtype: void Do not return anything, modify rooms in-place instead. """
if len(rooms)==0 or len(rooms[0])==0:
return
zero_list = []
for i in range(len(rooms)):
for j in range(len(rooms[0])):
if rooms[i][j]==0:
self.traverse4zero( (i,j), rooms )
def traverse4zero(self, zero_item, rooms):
bfs_list = [zero_item]
dict_direct = [(-1,0), (1,0), (0,1), (0,-1)]
while bfs_list!=[]:
cur_item = bfs_list[0]
bfs_list = bfs_list[1:]
for i in range(4):
next_i = cur_item[0] + dict_direct[i][0]
next_j = cur_item[1] + dict_direct[i][1]
if next_i<0 or next_i>=len(rooms) \
or next_j<0 or next_j>=len(rooms[0]) \
or rooms[next_i][next_j]<=0 \
or rooms[cur_item[0]][cur_item[1]]+1 >= rooms[next_i][next_j]:
continue
rooms[next_i][next_j] = min( rooms[next_i][next_j], \
rooms[cur_item[0]][cur_item[1]]+1 )
bfs_list.append( (next_i, next_j) )
returnSolution2
Accept
Time: 588ms
This solution is different from solution1. BFS is done for each zero element at the same time.
class Solution(object):
def wallsAndGates(self, rooms):
""" :type rooms: List[List[int]] :rtype: void Do not return anything, modify rooms in-place instead. """
if len(rooms)==0 or len(rooms[0])==0:
return
bfs_list = []
for i in range(len(rooms)):
for j in range(len(rooms[0])):
if rooms[i][j]==0:
bfs_list.append( (i,j) )
dict_direct = [(-1,0), (1,0), (0,1), (0,-1)]
while bfs_list!=[]:
cur_item = bfs_list[0]
del bfs_list[0]
for i in range(4):
next_i = cur_item[0] + dict_direct[i][0]
next_j = cur_item[1] + dict_direct[i][1]
if next_i<0 or next_i>=len(rooms) \
or next_j<0 or next_j>=len(rooms[0]) \
or rooms[cur_item[0]][cur_item[1]]+1 >= rooms[next_i][next_j]:
continue
rooms[next_i][next_j] = rooms[cur_item[0]][cur_item[1]]+1
bfs_list.append( (next_i, next_j) )
return Take home message
For popping the first element of list, use
del bfs_list[0]
instead of
bfs_list = bfs_list[1:]