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。

 1 class Solution {
 2     public int slidingPuzzle(int[][] board) {
 3         String target = "123450";
 4         String start = "";
 5         for (int i = 0; i < board.length; i++) {
 6             for (int j = 0; j < board[0].length; j++) {
 7                 start += board[i][j];
 8             }
 9         }
10         HashSet<String> visited = new HashSet<>();
11         // all the positions 0 can be swapped to
12         int[][] dirs = new int[][] { { 1, 3 }, { 0, 2, 4 },
13                 { 1, 5 }, { 0, 4 }, { 1, 3, 5 }, { 2, 4 } };
14         Queue<String> queue = new LinkedList<>();
15         queue.offer(start);
16         visited.add(start);
17         int res = 0;
18         while (!queue.isEmpty()) {
19             // level count, has to use size control here, otherwise not needed
20             int size = queue.size();
21             for (int i = 0; i < size; i++) {
22                 String cur = queue.poll();
23                 if (cur.equals(target)) {
24                     return res;
25                 }
26                 int zero = cur.indexOf('0');
27                 // swap if possible
28                 for (int dir : dirs[zero]) {
29                     String next = swap(cur, zero, dir);
30                     if (visited.contains(next)) {
31                         continue;
32                     }
33                     visited.add(next);
34                     queue.offer(next);
35 
36                 }
37             }
38             res++;
39         }
40         return -1;
41     }
42 
43     private String swap(String str, int i, int j) {
44         StringBuilder sb = new StringBuilder(str);
45         sb.setCharAt(i, str.charAt(j));
46         sb.setCharAt(j, str.charAt(i));
47         return sb.toString();
48     }
49 }