题目描述:
You have a lock in front of you with 4 circular wheels. Each wheel has 10 slots: '0', '1', '2', '3', '4', '5', '6', '7', '8', '9'
. The wheels can rotate freely and wrap around: for example we can turn '9'
to be '0'
, or '0'
to be '9'
. Each move consists of turning one wheel one slot.
The lock initially starts at '0000'
, a string representing the state of the 4 wheels.
You are given a list of deadends
dead ends, meaning if the lock displays any of these codes, the wheels of the lock will stop turning and you will be unable to open it.
Given a target
representing the value of the wheels that will unlock the lock, return the minimum total number of turns required to open the lock, or -1 if it is impossible.
Example 1:
Input: deadends = ["0201","0101","0102","1212","2002"], target = "0202" Output: 6 Explanation: A sequence of valid moves would be "0000" -> "1000" -> "1100" -> "1200" -> "1201" -> "1202" -> "0202". Note that a sequence like "0000" -> "0001" -> "0002" -> "0102" -> "0202" would be invalid, because the wheels of the lock become stuck after the display becomes the dead end "0102".
Example 2:
Input: deadends = ["8888"], target = "0009" Output: 1 Explanation: We can turn the last wheel in reverse to move from "0000" -> "0009".
Example 3:
Input: deadends = ["8887","8889","8878","8898","8788","8988","7888","9888"], target = "8888" Output: -1 Explanation: We can't reach the target without getting stuck.
Example 4:
Input: deadends = ["0000"], target = "8888" Output: -1
Note:
- The length of
deadends
will be in the range[1, 500]
. target
will not be in the listdeadends
.- Every string in
deadends
and the stringtarget
will be a string of 4 digits from the 10,000 possibilities'0000'
to'9999'
.
题目大意:
4位密码锁初始状态为0000,有一些状态不可以到达,记为deadends,求到达目标状态target的最少旋转次数。若target不可达,则返回-1。
解题思路:
BFS(广度优先搜索)
Python代码:
class Solution(object):
def openLock(self, deadends, target):
"""
:type deadends: List[str]
:type target: str
:rtype: int
"""
if "0000" in deadends: return -1
queue = ["0000"]
visits = set(queue + deadends)
step = 0
while queue:
queue0 = []
for status in queue:
if status == target: return step
for nstatus in self.nextStatus(status):
if nstatus in visits: continue
visits.add(nstatus)
queue0.append(nstatus)
queue = queue0
step += 1
return -1
def nextStatus(self, status):
ans = []
for i, n in enumerate(status):
for y in (-1, 1):
digit = (int(status[i]) + y + 10) % 10
yield status[:i] + str(digit) + status[i+1:]
本文链接:http://bookshadow.com/weblog/2017/12/24/leetcode-open-the-lock/
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