752. Open the Lock

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752. Open the Lock

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:

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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:

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Input: deadends = ["8888"], target = "0009"
Output: 1
Explanation:
We can turn the last wheel in reverse to move from "0000" -> "0009".

Example 3:

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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:

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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 list deadends.
  • 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

密码锁解锁，从 0000 开始，移动每个齿轮，直到匹配到目标数字组合，求需要移动的步数。
其中 deadends 表示不能够访问的数字组合； 0 可以转到 9, 9 也可以转到 0。

解题思路

BFS

代码

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class Solution {
    func openLock(_ deadends: [String], _ target: String) -> Int {
        if deadends.contains("0000") { return -1 }
        var queue = ["0000"]
        var visited = Set(deadends + queue)
        var step = 1
        while !queue.isEmpty {
            var temp = [String]()
            for status in queue {
                let chars = status.utf8
                for (i, char) in chars.enumerated() {
                    let pre = String(chars.prefix(i))!
                    let suf = String(chars.suffix(4-i-1))!
                    for j in [1, 9] {   //需要即能正向转动齿轮，亦可反向转动齿轮。
                        let e = (Int(char) - 48 + j) % 10
                        let str = pre  + "\(e)" + suf
                        if target == str { return step }
                        if !visited.contains(str) {
                            visited.insert(str)
                            temp.append(str)
                        }
                    }
                }
            }
            queue = temp
            step += 1
        }
        return -1
    }
}
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