LWC 69: 773. Sliding Puzzle

传送门：773. Sliding Puzzle

Problem:

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.
boardi will be a permutation of 0, 1, 2, 3, 4, 5.

思路：

状态数很少，BFS套路题，把[1, 2, 3,4, 5, 0]hash成字符串”123450”即可。

Java版本：

代码语言：javascript复制

    public int slidingPuzzle(int[][] board) {
        int n = board.length;
        int m = board[0].length;
        String ter = "123450";

        Queue<int[][]> q = new ArrayDeque<>();
        Set<String> vis = new HashSet<>();
        q.offer(board);
        vis.add(hash(board));

        int turn = 0;
        int[][] dir = {{1, 0},{-1, 0},{0, 1},{0, -1}};
        while (!q.isEmpty()) {
            int size = q.size();
            for (int i = 0; i < size;   i) {
                int[][] now = q.poll();
                if (hash(now).equals(ter)) {
                    return turn;
                }
                // find 0
                int x = -1;
                int y = -1;
                for (int k = 0; k < n;   k) {
                    for (int l = 0; l < m;   l) {
                        if (now[k][l] == 0) {
                            x = k;
                            y = l;
                            break;
                        }
                    }
                }

                for (int[] d : dir) {
                    int nx = d[0]   x;
                    int ny = d[1]   y;
                    if (nx >= 0 && nx < n && ny >= 0 && ny < m) {
                        int[][] aux = clone(now);
                        swap(aux, x, y, nx, ny);
                        if (!vis.contains(hash(aux))) {
                            q.offer(aux);
                            vis.add(hash(aux));
                        }

                    }
                }
            }
            turn   ;
        }
        return -1;
    }

    int[][] clone(int[][] board) {
        int n = board.length;
        int m = board[0].length;
        int[][] aux = new int[n][m];
        for (int i = 0; i < n;   i) {
            for (int j = 0; j < m;   j) {
                aux[i][j] = board[i][j];
            }
        }
        return aux;
    }

    String hash(int[][] board) {
        StringBuilder sb = new StringBuilder();
        int n = board.length;
        int m = board[0].length;
        for (int i = 0; i < n;   i) {
            for (int j = 0; j < m;   j) {
                sb.append(board[i][j]);
            }
        }
        return sb.toString();
    }

    void swap(int[][] board, int ai, int aj, int bi, int bj) {
        int tmp = board[ai][aj];
        board[ai][aj] = board[bi][bj];
        board[bi][bj] = tmp;
    }

Python版本：

代码语言：javascript复制

class Solution(object):
    def slidingPuzzle(self, board):
        """
        :type board: List[List[int]]
        :rtype: int
        """
        seen = set()
        END = '123450'
        INIT = ''.join(map(str, board[0]))   ''.join(map(str, board[1]))
        queue = []
        seen.add(INIT)
        queue.append(INIT)
        turn = 0
        while queue:
            size = len(queue)
            for i in range(size):
                now = queue.pop(0)
                if now == END: return turn
                idx = now.index('0')
                r = idx / 3
                c = idx % 3
                for d in (1, -1):
                    for nx, ny in ((r   d, c), (r, c   d)):      
                        if nx >= 0 and nx < 2 and ny >= 0 and ny < 3:
                            nidx = nx * 3   ny
                            nxt = list(now)
                            nxt[idx], nxt[nidx] = nxt[nidx], nxt[idx]
                            nxt = ''.join(nxt)
                            if nxt not in seen:
                                seen.add(nxt)
                                queue.append(nxt)
            turn  = 1
        return -1

python input output return state

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