题目描述：

LeetCode 773. Sliding Puzzle

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].

题目大意：

模拟实现滑块拼图游戏

给定2行3列的矩阵board，包含数字0 - 5，求将其恢复为[[1,2,3],[4,5,0]]的状态最少需要移动多少次。

解题思路：

广度优先搜索（BFS）

Python代码：

class Solution(object):
    def slidingPuzzle(self, board):
        """
        :type board: List[List[int]]
        :rtype: int
        """
        step = 0
        board = tuple(map(tuple, board))
        q = [board]
        memo = set([board])
        while q:
            q0 = []
            for b in q:
                if b == ((1,2,3), (4,5,0)): return step
                for x in range(2):
                    for y in range(3):
                        if b[x][y]: continue
                        for dx, dy in zip((1, 0, -1, 0), (0, 1, 0, -1)):
                            nx, ny = x + dx, y + dy
                            if 0 <= nx < 2 and 0 <= ny < 3:
                                nb = list(map(list, b))
                                nb[nx][ny], nb[x][y] = nb[x][y], nb[nx][ny]
                                nb = tuple(map(tuple, nb))
                                if nb not in memo:
                                    memo.add(nb)
                                    q0.append(nb)
            q = q0
            step += 1
        return -1

 

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