LeetCode Sudoku Solver 回溯

思路：

DFS。时间复杂度O(9^4)，空间复杂度O(1)。
下棋添子，下完一步下另一步，不行就回溯。

class Solution {private:    bool isValid(vector<vector<char>> &board, int i, int j) {        //check col        for(int row = 0; row < 9; ++row) {            if(row != i && board[row][j] == board[i][j]) {                return false;            }        }        //check row        for(int col = 0; col < 9; ++col) {            if(col != j && board[i][col] == board[i][j]) {                return false;            }        }        //check cube        for(int row = i - i%3; row < i + (3 - i%3); ++row) {            for(int col = j - j%3; col < j + (3 - j%3); ++col) {                if(row == i && col == j) continue;                if(board[row][col] == board[i][j]) {                    return false;                }            }        }        return true;    }    bool dfs(vector<vector<char>>& board) {        for(int i = 0; i < 9; ++i) {            for(int j = 0; j < 9; ++j) {                if(board[i][j] == '.') {                    for(int k = 1; k <= 9; ++k) {                        board[i][j] = '0' + k;                        if(isValid(board, i, j) && dfs(board)) {                            return true;                        }                        board[i][j] = '.';                    }                    return false;                }            }        }        return true;    }public:    void solveSudoku(vector<vector<char>>& board) {        dfs(board);    }};