【LeetCode 773】 Sliding Puzzle

题目描述

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].

思路

关键是要记录状态吧，我转换成string表示board的状态。
BFS。

代码

class Solution {
public:
    int slidingPuzzle(vector<vector<int>>& board) {
        if (check(board)) return 0;
        queue<string> que;
        string st = btos(board);
        que.push(st);
        vis[st] = 1;
        int step = 0;
        while(!que.empty()) {
            int size = que.size();
            while(size--) {
                string cur = que.front();
                que.pop();
                int index = 0;
                int stx, sty;
                for (int i=0; i<2; ++i) {
                    for (int j=0; j<3; ++j) {
                        board[i][j] = cur[index++] - '0';
                        if (board[i][j] == 0) {
                            stx = i, sty = j;
                        }
                    }
                }
            
                for (int i=0; i<4; ++i) {
                    int nx = stx + dir[i];
                    int ny = sty + dir[i+1];
                    if (nx < 0 || nx >= 2 || ny < 0 || ny >= 3) continue;
                    int tmp = board[nx][ny];
                    board[nx][ny] = 0;
                    board[stx][sty] = tmp;
                    if (check(board)) return step+1;
                    string tt = btos(board);
                    if (vis[tt]) {
                        board[nx][ny] = tmp;
                        board[stx][sty] = 0;
                        continue;
                    }
                    vis[tt] = 1;
                    que.push(tt);
                    board[nx][ny] = tmp;
                    board[stx][sty] = 0;
                }
            }
            step++;
        }
        return -1;
    }
    
private:
    map<string, int> vis;
    vector<int> dir {1, 0, -1, 0, 1};
    string btos(vector<vector<int>>& board) {
        string st = "";
        for (int i=0; i<2; ++i) {
            for (int j=0; j<3; ++j) {
                st += (board[i][j] + '0');
            }
        }
        return st;
    }
    
    bool check(vector<vector<int>>& board) {
        int st = 1;
        for (int i=0; i<2; ++i) {
            for (int j=0; j<3; ++j) {
                if (i == 1 && j == 2) {
                    if (board[i][j] != 0) return false;
                }else if (board[i][j] != st++) return false;
            }
        }
        return true;
    }
};

早上七点多起来晕晕乎乎到了晚上七点。。。
时光飞逝，岁月如梭啊。。。
希望姐姐婚礼事事顺利。