leetcode专题训练 37. Sudoku Solver

类似于八皇后问题的回溯法。时间复杂度爆炸，然而能过。我还以为过不了。
这段代码中的精髓段如下。判断是否能将数字i填进去，如果能填就试着填一下。如果成立，就返回1，如果不成立，就把之前用来试着填的状态全部改回初始状态。

if (row[r][i] || col[c][i] || box[boxn][i]) continue;
row[r][i] = 1;
col[c][i] = 1;
box[boxn][i] = 1;
board[r][c] = '0' + i;
if (solve(now + 1, board, le - 1) == 1) {
    return 1;
}
row[r][i] = 0;
col[c][i] = 0;
box[boxn][i] = 0;
board[r][c] = '.';

还需要注意，用了一个vector存储所有空的块的位置，就不用再挨个扫矩阵中的每个元素了，可以直接跳过非空的块。
综上，这道题的解题代码如下：

class Solution {
private:
    int row[10][11], col[10][11], box[10][11];
    struct Node{
        int r, c, boxn;
        Node(){}
        Node(int rr, int cc, int nn) {
            r = rr;
            c = cc;
            boxn = nn;
        }
    };
    vector<Node> emp; //存储所有空白块的索引
    int FindBox(int r, int c)
    {
        int tmp_r = r / 3;
        int tmp_c = c / 3;
        if (tmp_r == 0) {
            if (tmp_c == 0) return 0;
            if (tmp_c == 1) return 1;
            if (tmp_c == 2) return 2;
        } 
        if (tmp_r == 1) {
            if (tmp_c == 0) return 3;
            if (tmp_c == 1) return 4;
            if (tmp_c == 2) return 5;
        }
        if (tmp_r == 2) {
            if (tmp_c == 0) return 6;
            if (tmp_c == 1) return 7;
            if (tmp_c == 2) return 8;
        }
        return -1;
    }
    int solve(int now, vector<vector<char>>& board, int le)
    {
        if (le == 0) return 1;
        int r = emp[now].r;
        int c = emp[now].c;
        int boxn = emp[now].boxn;
        for (int i = 1; i <= 9; i++) {
            if (row[r][i] || col[c][i] || box[boxn][i]) continue;
            row[r][i] = 1;
            col[c][i] = 1;
            box[boxn][i] = 1;
            board[r][c] = '0' + i;
            if (solve(now + 1, board, le - 1) == 1) {
                return 1;
            }
            row[r][i] = 0;
            col[c][i] = 0;
            box[boxn][i] = 0;
            board[r][c] = '.';
        }
        return 0;
    }
public:
    void solveSudoku(vector<vector<char>>& board) {
        emp.clear();
        memset(row, 0, sizeof(row));
        memset(col, 0, sizeof(col));
        memset(box, 0, sizeof(box));
        int le = 0;
        for (int i = 0; i < 9; i++) {
            for (int j = 0; j < 9; j++) {
                if (board[i][j] == '.') {
                    le++;
                    emp.push_back(Node(i, j, FindBox(i, j)));
                    continue;
                }
                int num = board[i][j] - '0';
                row[i][num] = 1;
                col[j][num] = 1;
                box[FindBox(i, j)][num] = 1;
            }
        }
        solve(0, board, le);
        return;
    }
};