leetcode 752. Open the Lock

原题传送门->leetcode152 Open the Lock

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

Constraints:

1 <= deadends.length <= 500
deadends[i].length == 4
target.length == 4
target will not be in the list deadends.
target and deadends[i] consist of digits only.

入门级BFS。
但我水平还欠火候。
python也还不够熟。

from queue import Queue
class Solution:
    def __init__(self):
        self.visited = []
        self.tar = 0
        self.dead = []
    def openLock(self, deadends, target) -> int:
        if target=='0':
            return 0
        self.visited = [0 for i in range(10000)]
        self.tar = int(target)
        for end in deadends:
            self.visited[int(end)] = 1
        if self.visited[0]==1:
            return -1 #报错点3
        return self.bfs() # 报错点2

    def change(self, num):
        nextcode = []
        for i in range(4):
            tmp = (num // (10**i)) % 10 # 报错点1
            nextcode.append((10 ** i)*(((tmp + 1) % 10) - tmp) + num)
            nextcode.append((10 ** i)*(((tmp - 1) % 10) - tmp) + num)
        return tuple(nextcode)
    def bfs(self):
        q = Queue(maxsize=0)
        time = 0
        q.put(0)
        while(q.empty()==False):
            size = q.qsize()
            for i in range(size):
                tmp = q.get()
                if tmp == self.tar:
                    return time
                nextcode = self.change(tmp)
                for ii in nextcode:
                    if self.visited[ii] == 0:
                        q.put(ii)
                        self.visited[ii] = 1
            time += 1
        return -1

报错点:
1、”/“与”//“没分清
2、return写到循环里边了。。(应该写到外面)
3、deadends包含”0000”的情况没考虑