题目如下：
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].
解题思路：对于这个题目，我也没想到特别好的方法。不过既然题目约定了是一个2*3的board，那么基本上就不用考虑性能问题了，所以可以简单粗暴的用穷举法。怎么穷举呢，最简单的是倒推，因为如果题目有解的话最终的结果一定是 [[1,2,3],[4,5,0]]，我们可以用这个状态作为起点，计算出多少次移动能移动到和输入board一样的状态。因为只能移动0，而且只有上下左右四种移动方向，这个就是一个很典型的广度遍历的场景，注意每次移动后都要记录当前的状态，也要记录到达这个状态需要移动的次数，用来和board比较，如果一致就不需要再继续移动了。最后，当然也可以事先把所有能移动到达的状态先计算出来并进行缓存，之后就直接和board进行比较就行了。
完整代码：
import copy class Solution(object): def slidingPuzzle(self, board): """ :type board: List[List[int]] :rtype: int """ pass def toString(self,b): s = ‘‘ for i in b: for j in i: s += str(j) return s def getIndex(self,b): for i in range(len(b)): for j in range(len(b[0])): if b[i][j] == 0: return [i,j] return [] def swap(self,b,x1,y1,x2,y2): t = b[x1][y1] b[x1][y1] = b[x2][y2] b[x2][y2] = t def slidingPuzzle(self, board): des = [[1,2,3],[4,5,0]] #print self.toString(board) res = [] res.append(self.toString(des)) stack = [] stack.append([des,0]) while len(stack) > 0: e = stack[0] if e[0] == board: return e[1] #break del stack[0] #print ‘step‘,e[1] #print e[0][0] #print e[0][1] #print ‘**********‘ inx = self.getIndex(e[0]) #up: if inx[0] - 1 == 0: bup = copy.deepcopy(e[0]) self.swap(bup, inx[0], inx[1], inx[0] -1,inx[1]) if self.toString(bup) not in res: res.append(self.toString(bup)) stack.append([bup,e[1]+1]) #down if inx[0] + 1 == 1: bdown = copy.deepcopy(e[0]) self.swap(bdown, inx[0], inx[1], inx[0] + 1,inx[1]) if self.toString(bdown) not in res: res.append(self.toString(bdown)) stack.append([bdown,e[1]+1]) #left if inx[1] - 1 >= 0: bleft = copy.deepcopy(e[0]) self.swap(bleft, inx[0], inx[1], inx[0],inx[1]-1) if self.toString(bleft) not in res: res.append(self.toString(bleft)) stack.append([bleft,e[1]+1]) #right if inx[1] + 1 <= 2: bright = copy.deepcopy(e[0]) self.swap(bright, inx[0], inx[1], inx[0] ,inx[1]+1) if self.toString(bright) not in res: res.append(self.toString(bright)) stack.append([bright,e[1]+1]) return -1
