317 . Shortest Distance from All Buildings
典型的BFS的题目,BFS可以求得最短距离,所以很多碰到最小路径的时候可以考虑用BFS,还有这道题可以在访问的时候多记录一些信息,可以达到剪枝的效果
class Solution(object):
def shortestDistance(self, grid):
"""
:type grid: List[List[int]]
:rtype: int
"""
m = {} # building (i, j) -> {} # 0 x,y -> distance to i,j
count = 0
total = 0
for i in range(len(grid)):
for j in range(len(grid[0])):
if grid[i][j] == 1:
total += 1
for i in range(len(grid)):
for j in range(len(grid[0])):
if grid[i][j] == 1:
count += 1
building = (i, j)
m[building] = {}
if not self.bfs(grid, building, m, total):
return -1
h = {}
for key, val in m.iteritems():
if not val:
return -1
for k, v in val.iteritems():
if k in h:
h[k] = [h[k][0]+v, h[k][1]+1] # total steps, number of buildings reached
else:
h[k] = [v, 1]
if not h:
return -1
res = sys.maxint
found = False
for v in h.values():
if v[1] == count:
res = min(res, v[0])
found = True
if found:
return res
else:
return -1
def bfs(self, grid, building, m, total):
dist = 0
i, j = building
count_building = 1
visited = [[False for _ in range(len(grid[0]))] for _ in range(len(grid))]
visited[i][j] = True
queue = [[i+1, j], [i-1, j], [i, j+1], [i, j-1]]
while queue:
l = len(queue)
dist += 1
for k in range(l):
x, y = queue.pop(0)
if 0 <= x < len(grid) and 0 <= y < len(grid[0]) and not visited[x][y]:
visited[x][y] = True
if grid[x][y] == 0:
if (x, y) in m[building]:
continue
else:
m[building][(x,y)] = dist
queue += [[x+1, y], [x-1, y], [x, y+1], [x, y-1]]
elif grid[x][y] == 1:
count_building += 1
return count_building == total
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