- Series: Sliding Puzzle & Follow-up
- LintCode: Sliding Puzzle III
- Review: Game Problems
- Tag: #graph, #game, #slidingpuzzle
On a 2×3 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.
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
- 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].
Credits To: leetcode.com
Leave me comments, if you have better ways to solve.
## Blog link: https://code.dennyzhang.com/sliding-puzzle ## Basic Ideas: BFS: find the mininum path from point1 to point2 ## ## Complexity: class Solution(object): def slidingPuzzle(self, board): """ :type board: List[List[int]] :rtype: int """ target = '123450' queue, visited = , set() queue.append(self.toString(board)) level = 0 while len(queue) != 0: for i in range(len(queue)): state = queue if state == target: return level del queue visited.add(state) # move to next self.addNeighbors(state, visited, queue) # print queue level += 1 return -1 def toString(self, board): res = '' for i in range(2): for j in range(3): res = res + str(board[i][j]) return res def addNeighbors(self, state, visited, queue): # change back state to board matrix = [[None for j in range(3)] for i in range(2)] index = 0 i0, j0 = None, None for i in range(2): for j in range(3): matrix[i][j] = state[index] index += 1 if matrix[i][j] == '0': i0, j0 = i, j # get next: for (ik, jk) in [(-1, 0), (1, 0), (0, 1), (0, -1)]: i2, j2 = i0+ik, j0+jk if i2<0 or i2 >= 2 or j2<0 or j2>=3: continue matrix[i0][j0], matrix[i2][j2] = matrix[i2][j2], matrix[i0][j0] newState = self.toString(matrix) if newState not in visited: queue.append(newState) # change back matrix[i0][j0], matrix[i2][j2] = matrix[i2][j2], matrix[i0][j0]