数独

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.

分析：暴力枚举，dfs遍历每个格子，依次从1到9放数，如果合法继续，否则返回。

#include <iostream>using namespace std;char board[9][9];// LeetCode, Sudoku Solver// 时间复杂度O(9^4)，空间复杂度O(1)class Solution {public:    bool solveSudoku(char board[][9]) {        for (int i = 0; i < 9; ++i)            for (int j = 0; j < 9; ++j) {                if (board[i][j] == '.') {                    for (int k = 0; k < 9; ++k) {                        board[i][j] = '1' + k;                        if (isValid(board, i, j) && solveSudoku(board))                            return true;                        board[i][j] = '.';                    }                    return false;                }            }        return true;    }private:    // 检查 (x, y) 是否合法    bool isValid(char board[][9], int x, int y) {        int i, j;        for (i = 0; i < 9; i++) // 检查 y 列            if (i != x && board[i][y] == board[x][y])                return false;        for (j = 0; j < 9; j++) // 检查 x 行            if (j != y && board[x][j] == board[x][y])                return false;        for (i = 3 * (x / 3); i < 3 * (x / 3 + 1); i++)            for (j = 3 * (y / 3); j < 3 * (y / 3 + 1); j++)                if ((i != x || j != y) && board[i][j] == board[x][y])                    return false;        return true;    }};int main(){int T;cin >> T;while(T--) {for(int i=0; i<9; i++)for(int j=0; j<9; j++)cin >> board[i][j];Solution s;bool ans = s.solveSudoku(board);if (ans) {for(int i=0; i<9; i++) {for(int j=0; j<9; j++)cout << board[i][j] << " ";cout << endl;}} elsecout << "no solution" << endl;cout << endl;}    return 0;}