[LeetCode] Sudoku Solver

Write a program to solve a Sudoku puzzle by filling the empty cells.

Empty cells are indicated by the character '.'.

You may assume that there will be only one unique solution.

A sudoku puzzle...

...and its solution numbers marked in red.

终于到了朝思暮想的sudoku solver了， 有了Valid Sudoku的基础，我知道了用哈希表可以很方便的判断冲突。思路还是遍历这个board，这次是找到为空的位置，然后对可能的结果集进行筛选：把行、列、3x3出现的数字都去掉，剩下的就是可能的结果集。然后把结果集里的元素一个一个拿出来进行深度搜索，直到满足递归出口：row == 8 && col == 8 说明board遍历完毕，得到结果； 如果函数最终返回却没有满足那个条件的话，就说明此局无解。

在解此题时主要遇到的问题是对递归出口判断的模棱两可，还有传board的时候我老是喜欢传个引用，导致一次深搜改变了board。。这是个bug，在word breakII时也遇到过。总值对此题的理解还需深入，继续推敲。

void solver(vector<vector<char>> board, int row, int col, vector<vector<char>> &ret) {
    if (board[row][col] != '.') {
        if (row == 8 && col == 8) {
            ret = board;
            return;
        }
        else if (++col >= 9) {
            ++row;
            col = 0;
        }
        solver(board, row, col, ret);
        return;
    }
    
    // calculate solve space
    unordered_set<char> set = preset;
    // filter colum
    for (int k = 0; k < 9; k++) {
        char searched = board[k][col];
        if (searched != '.' && set.find(searched) != set.end()) {
            set.erase(searched);
        }
    }
    // filter row
    for (int k = 0; k < 9; k++) {
        char searched = board[row][k];
        if (searched != '.' && set.find(searched) != set.end()) {
            set.erase(searched);
        }
    }
    // filter 3*3
    int square_row_offset = row / 3;
    int square_col_offset = col / 3;
    for (int k = square_row_offset * 3; k < square_row_offset * 3 + 3; k++) {
        for (int l = square_col_offset * 3; l < square_col_offset * 3 + 3; l++) {
            char searched = board[k][l];
            if (searched != '.' && set.find(searched) != set.end()) {
                set.erase(searched);
            }
        }
    }
    
    for (char ele: set) {
        board[row][col] = ele;
        
        if (row == 8 && col == 8) {
            ret = board;
            return;
        }
        
        solver(board, row, col+1, ret);
    }
}


vector<vector<char>> sudokuSolver(vector<vector<char>> &board) {
    vector<vector<char>> ret;
    solver(board, 0, 0, ret);
    return ret;
}