752. Open the Lock
package LeetCode_752 import java.util.* import kotlin.collections.HashSet /** * 752. Open the Lock * https://leetcode.com/problems/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". * */ class Solution { /* * solution: BFS, find the shortest path in weighted undirected graph, * "0000" as a node, each node has 8 edge, because each number can roll up or roll down, 4*2=8, each weight is 1; * Time complexity: O(8*10000), because 0000-9999 has 10000 status; * Space complexity: O(10000) * */ fun openLock(deadends: Array<String>, target: String): Int { //put deadends into set, can improve search speed val set = HashSet<String>() for (item in deadends) { set.add(item) } val start = "0000" if (set.contains(start)) { return -1 } if (target == start){ return 0 } val queue = LinkedList<String>() val visited = HashSet<String>() queue.offer(start) visited.add(start) var step = 0 while (queue.isNotEmpty()) { step++ //size represent the count of node in current level val size = queue.size //expand the node for (i in 0 until size){ val cur = queue.pop() //there are 4 number in each node for (n in 0 until 4){ val c = cur[n] //simulate roll up or roll down val str1 = cur.substring(0,n) + (if (c=='9') 0 else c-'0'+1) + cur.substring(n+1) val str2 = cur.substring(0,n) + (if (c=='0') 9 else c-'0'-1) + cur.substring(n+1) if (str1==target || str2==target){ return step } if (!visited.contains(str1) && !set.contains(str1)){ visited.add(str1) queue.offer(str1) } if (!visited.contains(str2) && !set.contains(str2)){ visited.add(str2) queue.offer(str2) } } } } //if not found solution return -1 } }
【推荐】国内首个AI IDE,深度理解中文开发场景,立即下载体验Trae
【推荐】编程新体验,更懂你的AI,立即体验豆包MarsCode编程助手
【推荐】抖音旗下AI助手豆包,你的智能百科全书,全免费不限次数
【推荐】轻量又高性能的 SSH 工具 IShell:AI 加持,快人一步
· AI与.NET技术实操系列(二):开始使用ML.NET
· 记一次.NET内存居高不下排查解决与启示
· 探究高空视频全景AR技术的实现原理
· 理解Rust引用及其生命周期标识(上)
· 浏览器原生「磁吸」效果!Anchor Positioning 锚点定位神器解析
· DeepSeek 开源周回顾「GitHub 热点速览」
· 物流快递公司核心技术能力-地址解析分单基础技术分享
· .NET 10首个预览版发布:重大改进与新特性概览!
· AI与.NET技术实操系列(二):开始使用ML.NET
· .NET10 - 预览版1新功能体验(一)