LeetCode 37. Sudoku Solver
Write a program to solve a Sudoku puzzle by filling the empty cells.
A sudoku solution must satisfy all of the following rules:
Each of the digits 1-9 must occur exactly once in each row.
Each of the digits 1-9 must occur exactly once in each column.
Each of the digits 1-9 must occur exactly once in each of the 9 3x3 sub-boxes of the grid.
The '.' character indicates empty cells.
Example 1:
Input: 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"]]
Output: [["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"]]
Explanation: The input board is shown above and the only valid solution is shown below:
Constraints:
board.length == 9
board[i].length == 9
board[i][j] is a digit or '.'.
It is guaranteed that the input board has only one solution.
实现思路：
这道题很有难度，比起N皇后问题要更加地复杂，本题采用两种方法去实现，一种采用双重循环+dfs回溯剪枝的方式，一种采用N-皇后问题的单重循环回溯算法的解决方式去做，先来思考一些N-皇后问题，如果是N-皇后问题中，9X9的棋盘格子中，因为固定了皇后都放在不同一行，只需要根据皇后放置在不同列的情况进行穷举，并且皇后这个数字是唯一的'Q'字符，所以采用了for循环中进行dfs的一个单循环回溯算法方式，但本题数独的问题中，这个数字变量是大大提高了的，首先数字可以放在不同行不同列，并且一个坐标上可以放的数字又是1-9，所以情况就变多了，这里第一种答题的回溯模式，就是三重循环+dfs回溯，本题需要注意只需要给出一组答案即可，剩余答案需要进行剪枝操作，题目要求是按照行优先从1-9循环给数求数独答案的方式，求出这一组解即可。
先给出回溯图的一小部分，这里引用leetcode题解代码随想录作者的图
AC代码：
方法一：
采用三重循环+dfs回溯的方式
class Solution {
bool judge(int row,int col,char c,vector<vector<char>> &board) {
for(int i=0; i<board.size(); i++)
if(board[row][i]==c) return false;//一行上是否有相同元素
for(int j=0; j<9; j++)
if(board[j][col]==c) return false;//一列上是否存在相同元素
//判断在一个小方格内是否有元素重复
int _row=(row/3)*3;//找出处于第几个大方格 9个9宫格
int _col=(col/3)*3;
for(int i=_row; i<_row+3; i++) {
for(int j=_col; j<_col+3; j++) {
if(board[i][j]==c) return false;//九宫格内是否有相同元素
}
}
return true;//可以放置元素
}
bool dfs(vector<vector<char>> &board) {
for(int x=0; x<9; x++) {
for(int y=0; y<9; y++) {
if(board[x][y]!='.') continue;
for(char u='1'; u<='9'; u++) {
if(judge(x,y,u,board)) {//是个合适的放元素的位置
board[x][y]=u;//放下元素
//找到合适的一组后直接要返回了 不需要继续找了 剪枝
if(dfs(board)) return true;
board[x][y]='.';//回溯
}
}
return false;//这句语句很关键 当1~9都不能放到当前小方格时不合法
}
}
return true;//说明最终返回了确认的位置
}
public:
void solveSudoku(vector<vector<char>>& board) {
dfs(board);
}
};
方法二：
办法二是将本题的棋盘抽象成类似N-皇后问题的解答方式，将每一个坐标(x,y)抽象成一个方格，用三个数组来判断行列和九宫格内数是否重复的方法，巧妙利用空间换时间，所以方法二的效率达到100%，远远比方法一要好。
class Solution {
private:
bool line[9][9]= {0};//利用数组存储某行存数的情况
bool column[9][9]= {0};
bool block[3][3][9]= {0};
bool valid=false;//当获得一个合法的数独解时为true
vector<pair<int, int>> spaces;//用来存放一个个格子的坐标(x,y)
public:
void dfs(vector<vector<char>>& board, int pos) {
if (pos == spaces.size()) {//当所有需要填写数字的格子都填了数字
valid = true;
return;
}
auto [i, j] = spaces[pos];
for (int digit = 0; digit < 9&&!valid ; ++digit) {//如果已经有解了就剪枝
if (!line[i][digit] && !column[j][digit] && !block[i / 3][j / 3][digit]) {//行列和九宫格中都没有元素重复
line[i][digit] = column[j][digit] = block[i / 3][j / 3][digit] = true;
board[i][j] = digit + '0' + 1;
dfs(board, pos + 1);
line[i][digit] = column[j][digit] = block[i / 3][j / 3][digit] = false;
}
}
}
void solveSudoku(vector<vector<char>>& board) {
for (int i = 0; i < 9; ++i) {
for (int j = 0; j < 9; ++j) {
if (board[i][j] == '.') {
spaces.push_back({i,j});
} else {
int digit = board[i][j] - '0' - 1;//因为真实数据和数组下标相差1
line[i][digit] = column[j][digit] = block[i / 3][j / 3][digit] = true;
}
}
}
dfs(board, 0);
}
};
