Quantcast
Channel: 书影 - Entries for the tag leetcode
Viewing all articles
Browse latest Browse all 559

[LeetCode]Open the Lock

$
0
0

题目描述:

LeetCode 752. 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: 6Explanation:
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: 1Explanation:
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: -1Explanation:
We can't reach the target without getting stuck.

Example 4:

Input: deadends = ["0000"], target = "8888"Output: -1

Note:

  1. The length of deadends will be in the range [1, 500].
  2. target will not be in the list deadends.
  3. Every string in deadends and the string target 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:]

 


Viewing all articles
Browse latest Browse all 559

Latest Images

Trending Articles



Latest Images