编写一个程序,通过填充空格来解决数独问题。
数独的解法需遵循如下规则:
- 数字
1-9
在每一行只能出现一次。 - 数字
1-9
在每一列只能出现一次。 - 数字
1-9
在每一个以粗实线分隔的3x3
宫内只能出现一次。(请参考示例图)
数独部分空格内已填入了数字,空白格用 '.'
表示。
示例 1:
输入:board = [["5","3",".",".","7",".",".",".","."],["6",".",".","1","9","5",".",".","."],[".","9","8",".",".",".",".","6","."],["8",".",".",".","6",".",".",".","3"],["4",".",".","8",".","3",".",".","1"],["7",".",".",".","2",".",".",".","6"],[".","6",".",".",".",".","2","8","."],[".",".",".","4","1","9",".",".","5"],[".",".",".",".","8",".",".","7","9"]]
输出:[["5","3","4","6","7","8","9","1","2"],["6","7","2","1","9","5","3","4","8"],["1","9","8","3","4","2","5","6","7"],["8","5","9","7","6","1","4","2","3"],["4","2","6","8","5","3","7","9","1"],["7","1","3","9","2","4","8","5","6"],["9","6","1","5","3","7","2","8","4"],["2","8","7","4","1","9","6","3","5"],["3","4","5","2","8","6","1","7","9"]]
解释:输入的数独如上图所示,唯一有效的解决方案如下所示:
提示:
board.length == 9
board[i].length == 9
board[i][j]
是一位数字或者'.'
- 题目数据 保证 输入数独仅有一个
cpp
class Solution {
public:
bool isvaild(int row,int col,char val,vector<vector<char>>& board){
//row
for(int i = 0;i < 9;i++){
if(board[row][i] == val) return false;
}
//col
for(int j = 0;j < 9;j++){
if(board[j][col] == val) return false;
}
//九宫格
int startx = (row/3)*3; // 假如在第一个九宫格,row/3=0,再*3=0;
int starty = (col/3)*3; //假如在第二个九宫格,row/3=1,再*3=3; 我直呼nb
for(int i = startx;i < startx+3;i++){
for(int j = starty;j < starty+3;j++){
if(board[i][j] == val) return false;
}
}
return true;
}
bool backtracking(vector<vector<char>>& board){
for(int i = 0;i < board.size();i++){
for(int j = 0;j < board[0].size();j++){
//遇到空格
if(board[i][j] == '.'){
for(char a = '1';a <= '9';a++){
//判断这里应该填入啥数字合法
if(isvaild(i,j,a,board)){
board[i][j] = a;
//得将这个状态一直返回
if(backtracking(board) == true) return true;
board[i][j] = '.'; // 回溯
}
}
return false; //填入0-9都不对,都不合法,填错了。
}
}
}
return true; //填完且填正确了。
}
void solveSudoku(vector<vector<char>>& board) {
backtracking(board);
}
};