【DFS + backtrack】LeetCode 37. Sudoku Solver

最新推荐文章于 2024-07-30 16:17:15 发布

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#LeetCode 37. Sudoku Solver

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本文介绍了解决LeetCode 37题“解数独”的一种方法。使用深度优先搜索(DFS)策略填充空缺位置，并通过剪枝减少搜索范围。实现中预设了已知数字的位置，避免无效尝试。

LeetCode 37. Sudoku Solver

Solution1：
转载自花花酱：http://zxi.mytechroad.com/blog/searching/leetcode-37-sudoku-solver/
与之前的DFS套路题不大一样，注意！

class Solution {
public:
    void solveSudoku(vector<vector<char>>& board) {
        rows_ = vector<vector<int>>(9, vector<int>(10));
        cols_ = vector<vector<int>>(9, vector<int>(10));
        boxes_ = vector<vector<int>>(9, vector<int>(10));

        //预先设置剪枝条件
        for(int i = 0; i < 9; i++)
            for(int j = 0; j < 9; j++) {
                const char c = board[i][j];                
                if ( c != '.') {
                    int n = c - '0';                    
                    int bx = i / 3;
                    int by = j / 3;
                    rows_[i][n] = 1;
                    cols_[j][n] = 1;
                    boxes_[bx * 3 + by][n] = 1;
                }
            }

        fill(board, 0, 0);
    }

private:
    vector<vector<int>> rows_, cols_, boxes_;

    bool fill(vector<vector<char>>& board, int x, int y) {
        if (x == 9)
            return true;

        int ny = (y + 1) % 9;
        int nx = (ny == 0) ? x + 1 : x;

        if (board[x][y] != '.') return fill(board, nx, ny);

        for (int i = 1; i <= 9; i++) {
            int bx = x / 3;
            int by = y / 3;
            int box_key = bx * 3 + by;
            if (!rows_[x][i] && !cols_[y][i] && !boxes_[box_key][i]) {
                rows_[x][i] = 1;
                cols_[y][i] = 1;
                boxes_[box_key][i] = 1;
                board[x][y] = i + '0';
                if (fill(board, nx, ny)) return true;
                board[x][y] = '.';
                boxes_[box_key][i] = 0;
                cols_[y][i] = 0;
                rows_[x][i] = 0;
            }
        }
        return false;
    }
};