解数独-算法题

题意：给一个数独题目，将空格填满；
传送门

输入：board = [[“5”,“3”,".",".",“7”,".",".",".","."],[“6”,".",".",“1”,“9”,“5”,".",".","."],[".",“9”,“8”,".",".",".",".",“6”,"."],[“8”,".",".",".",“6”,".",".",".",“3”],[“4”,".",".",“8”,".",“3”,".",".",“1”],[“7”,".",".",".",“2”,".",".",".",“6”],[".",“6”,".",".",".",".",“2”,“8”,"."],[".",".",".",“4”,“1”,“9”,".",".",“5”],[".",".",".",".",“8”,".",".",“7”,“9”]]
输出：[[“5”,“3”,“4”,“6”,“7”,“8”,“9”,“1”,“2”],[“6”,“7”,“2”,“1”,“9”,“5”,“3”,“4”,“8”],[“1”,“9”,“8”,“3”,“4”,“2”,“5”,“6”,“7”],[“8”,“5”,“9”,“7”,“6”,“1”,“4”,“2”,“3”],[“4”,“2”,“6”,“8”,“5”,“3”,“7”,“9”,“1”],[“7”,“1”,“3”,“9”,“2”,“4”,“8”,“5”,“6”],[“9”,“6”,“1”,“5”,“3”,“7”,“2”,“8”,“4”],[“2”,“8”,“7”,“4”,“1”,“9”,“6”,“3”,“5”],[“3”,“4”,“5”,“2”,“8”,“6”,“1”,“7”,“9”]]

题解一：DFS+回溯

void solveSudoku(vector<vector<char>>& board) {
    
    
    vector<vector<int>> line(9, vector<int>(10));
    vector<vector<int>> column(9, vector<int>(10));
    vector<vector<vector<int>>> block(3, vector<vector<int>>(3, vector<int>(10)));
    vector<pair<int, int>> spaces;
    for (int i = 0; i < 9; ++i) {
    
    
        for (int j = 0; j < 9; ++j) {
    
    
            if (board[i][j] == '.')
                spaces.push_back({
    
    i, j});
            else {
    
    
                int num = board[i][j]-'0';
                line[i][num] = column[j][num] = block[i/3][j/3][num] = 1;
            }
        }
    }
    bool is_ok = 0;
    function<void(int)> dfs = [&](int pos) {
    
    
        if (pos == spaces.size()) {
    
    
            is_ok = true;
            return;
        }
        auto [i, j] = spaces[pos];
        for (int digit = 1; digit <= 9 && !is_ok; ++digit) {
    
    
            if (!line[i][digit] && !column[j][digit] && !block[i/3][j/3][digit]) {
    
    
                line[i][digit] = column[j][digit] = block[i/3][j/3][digit] = 1;
                board[i][j] = digit+'0';
                dfs(pos+1);
                line[i][digit] = column[j][digit] = block[i/3][j/3][digit] = 0;
            }
        }
    };
    dfs(0);
}

题解二：记录二进制优化

void solveSudoku(vector<vector<char>>& board) {
    
    
    vector<int> line(9), column(9);
    vector<vector<int>> block(3, vector<int>(3)); 
    
    function<void(int,int,int)> mark = [&](int i, int j, int digit) {
    
    
        line[i] ^= (1<<digit);
        column[j] ^= (1<<digit);
        block[i/3][j/3] ^= (1<<digit);
    };

    int full = 511;
    bool is_ok = false;
    vector<pair<int, int>> spaces;
    function<void(int)> dfs = [&](int pos) {
    
    
        if (pos == spaces.size()) {
    
    
            is_ok = true;
            return;
        }
        auto [i, j] = spaces[pos];
        for (int mask = ~(line[i]|column[j]|block[i/3][j/3])&full; mask && !is_ok; mask &= (mask-1)) {
    
    
        	// 保留mask最低位的1，其余置0
            int digitMask = mask&(-mask);
            // 返回该数字后面连续0的个数
            int digit = __builtin_ctz(digitMask);
            mark(i, j, digit);
            board[i][j] = digit + '0' + 1;
            dfs(pos+1);
            mark(i, j, digit);
        }
    };

    for (int i = 0; i < 9; ++i) {
    
    
        for (int j = 0; j < 9; ++j) {
    
    
            if (board[i][j] == '.')
                spaces.push_back({
    
    i, j});
            else {
    
    
                int digit = board[i][j] - '0' - 1;
                mark(i, j, digit);
            }
        }
    }

    dfs(0);
}