On a 2x3 board, there are 5 tiles represented by the integers 1 through 5, and an empty square represented by 0.

A move consists of choosing 0 and a 4-directionally adjacent number and swapping it.

The state of the board is solved if and only if the board is [[1,2,3],[4,5,0]].

Given a puzzle board, return the least number of moves required so that the state of the board is solved. If it is impossible for the state of the board to be solved, return -1.

Examples:

Input: board = [[1,2,3],[4,0,5]]
Output: 1
Explanation: Swap the 0 and the 5 in one move.
Input: board = [[1,2,3],[5,4,0]]
Output: -1
Explanation: No number of moves will make the board solved.
Input: board = [[4,1,2],[5,0,3]]
Output: 5
Explanation: 5 is the smallest number of moves that solves the board.
An example path:
After move 0: [[4,1,2],[5,0,3]]
After move 1: [[4,1,2],[0,5,3]]
After move 2: [[0,1,2],[4,5,3]]
After move 3: [[1,0,2],[4,5,3]]
After move 4: [[1,2,0],[4,5,3]]
After move 5: [[1,2,3],[4,5,0]]
Input: board = [[3,2,4],[1,5,0]]
Output: 14

Note:

  • board will be a 2 x 3 array as described above.
  • board[i][j] will be a permutation of [0, 1, 2, 3, 4, 5].

分析:这题可以用bfs解,但是刚拿到这题的时候,完全想不到这题是可以用bfs解决的。

 class Solution {
public int slidingPuzzle(int[][] board) {
String target = "";
String start = "";
for (int i = ; i < board.length; i++) {
for (int j = ; j < board[].length; j++) {
start += board[i][j];
}
}
HashSet<String> visited = new HashSet<>();
// all the positions 0 can be swapped to
int[][] dirs = new int[][] { { , }, { , , },
{ , }, { , }, { , , }, { , } };
Queue<String> queue = new LinkedList<>();
queue.offer(start);
visited.add(start);
int res = ;
while (!queue.isEmpty()) {
// level count, has to use size control here, otherwise not needed
int size = queue.size();
for (int i = ; i < size; i++) {
String cur = queue.poll();
if (cur.equals(target)) {
return res;
}
int zero = cur.indexOf('');
// swap if possible
for (int dir : dirs[zero]) {
String next = swap(cur, zero, dir);
if (visited.contains(next)) {
continue;
}
visited.add(next);
queue.offer(next); }
}
res++;
}
return -;
} private String swap(String str, int i, int j) {
StringBuilder sb = new StringBuilder(str);
sb.setCharAt(i, str.charAt(j));
sb.setCharAt(j, str.charAt(i));
return sb.toString();
}
}

Sliding Puzzle的更多相关文章

  1. leetcode 542. 01 Matrix 、663. Walls and Gates(lintcode) 、773. Sliding Puzzle 、803. Shortest Distance from All Buildings

    542. 01 Matrix https://www.cnblogs.com/grandyang/p/6602288.html 将所有的1置为INT_MAX,然后用所有的0去更新原本位置为1的值. 最 ...

  2. Leetcode之广度优先搜索(BFS)专题-773. 滑动谜题(Sliding Puzzle)

    Leetcode之广度优先搜索(BFS)专题-773. 滑动谜题(Sliding Puzzle) BFS入门详解:Leetcode之广度优先搜索(BFS)专题-429. N叉树的层序遍历(N-ary ...

  3. [Swift]LeetCode773. 滑动谜题 | Sliding Puzzle

    On a 2x3 board, there are 5 tiles represented by the integers 1 through 5, and an empty square repre ...

  4. [LeetCode] Sliding Puzzle 滑动拼图

    On a 2x3 board, there are 5 tiles represented by the integers 1 through 5, and an empty square repre ...

  5. LeetCode 773. Sliding Puzzle

    原题链接在这里:https://leetcode.com/problems/sliding-puzzle/description/ 题目: On a 2x3 board, there are 5 ti ...

  6. [LeetCode] 773. Sliding Puzzle 滑动拼图

    On a 2x3 board, there are 5 tiles represented by the integers 1 through 5, and an empty square repre ...

  7. 【leetcode】Sliding Puzzle

    题目如下: On a 2x3 board, there are 5 tiles represented by the integers 1 through 5, and an empty square ...

  8. 【LeetCode】773. Sliding Puzzle 解题报告(Python)

    作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 题目地址: https://leetcode.com/problems/sliding- ...

  9. 滑动拼图 Sliding Puzzle

    2018-09-09 22:01:02 问题描述: 问题求解: 问题很Interesting,其实本质就是解空间遍历,使用BFS就可以很快的予以解决~ public int slidingPuzzle ...

随机推荐

  1. PHP mysqli_get_client_stats() 函数

    定义和用法 mysqli_get_client_stats() 函数返回有关客户端每个进程的统计. 语法 mysqli_get_client_stats(); 返回有关客户端每个进程的统计: < ...

  2. learning gcc __BEGIN_DECLS and __END_DECLS

    __BEGIN_DECLS and  __END_DECLS  be use for mix C and C++

  3. EnumHelper.cs

    网上找的,还比较实用的: using System; using System.Collections.Generic; using System.ComponentModel; using Syst ...

  4. [Luogu] 次小生成树

    https://www.luogu.org/problemnew/show/P4180#sub 严格次小生成树,即不等于最小生成树中的边权之和最小的生成树 首先求出最小生成树,然后枚举所有不在最小生成 ...

  5. Flask-Response

    Flask中的HTTPResponse from flask import Flask,redirect app = Flask(__name__) @app.route("/index&q ...

  6. AcWing:237. 程序自动分析(离散化 + 并查集)

    在实现程序自动分析的过程中,常常需要判定一些约束条件是否能被同时满足. 考虑一个约束满足问题的简化版本:假设x1,x2,x3,…x1,x2,x3,…代表程序中出现的变量,给定n个形如xi=xjxi=x ...

  7. 安装完Pycharm,启动时碰到"failed to load jvm dll"的解决方案

    今天安装完系统,配置pycharm的环境的时候,启动pycharm时,碰到"failed to load jvm dll"的错误, 下面给出其解决方案: 安装Microsoft V ...

  8. 记录一次webpack3升级到webpack4过程

    升级之前也参考了一些网上的教程.借鉴之,进行的自己的升级.一些版本为什么设为那个版本号也是参考别人的结果. 整体是按照先升级npm run dev:在升级npm run build的顺序. 首先升级w ...

  9. mock的使用

    mock的重要性 mock就是对于某些不容易构造或者不容易获取的对象,用一个虚拟的对象来创建的方法.项目开发和测试过程中,遇到以下的情况时,就需要模拟结果返回. 1.当另一方接口或服务还未完成,阻碍项 ...

  10. Linux批量处理常用方式

    批量处理思路在工作中使用的频率比较高,比如批量清理进程.批量删除文件.批量机器执行脚本等. 一.批量清理带java字样的进程 方式1:使用shell while语法. ${line}; done sh ...