题目：原题链接（中等）
标签：广度优先搜索
解法 | 时间复杂度 | 空间复杂度 | 执行用时 |
Ans 1 (Python) | O ( 10000 ) O(10000) O(10000) | O ( 10000 ) O(10000) O(10000) | 568ms (76.79%) |
Ans 2 (Python) | |||
Ans 3 (Python) |
解法一：
class Solution:
def openLock(self, dead_ends: List[str], target: str) -> int:
def neighbors(ss):
res = []
for ii in range(4):
res.append(ss[:ii] + str((int(ss[ii]) - 1) % 10) + ss[ii + 1:])
res.append(ss[:ii] + str((int(ss[ii]) + 1) % 10) + ss[ii + 1:])
return res
dead_ends = set(dead_ends)
if "0000" in dead_ends or target in dead_ends:
return -1
if target == "0000":
return 0
visited = dead_ends | {target}
queue = collections.deque([target])
step = 0
while queue:
step += 1
for _ in range(len(queue)):
s1 = queue.popleft()
for s2 in neighbors(s1):
if s2 == "0000":
return step
if s2 not in visited:
visited.add(s2)
queue.append(s2)
return -1
