37. 解数独

题目描述

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解

解答

class Solution {
public:void solveSudoku(vector<vector<char>>& board) {// 将每行的每个位置都用1-9去尝试，回溯，类似于dfs的思路// 满足条件是同行同列及所在3x3格子内都不出现backtrack(board);}bool backtrack(vector<vector<char>>& board){// 按行搜索每一个位置的可能情况for(int i = 0; i < board.size(); i++){for(int j = 0; j < board[0].size(); j++){if(board[i][j] == '.'){for(char k = 1; k <= 9; k++){// 判断在数独棋盘(i, j) 点放k值是否合法if(isValid(i, j, k + '0', board)){board[i][j] = k + '0';if(backtrack(board)) return true;board[i][j] = '.';}}// 某一个格子放1-9都不行,返回false，从而能够回溯到上一层重新选取return false;}}}return true;}bool isValid(int row, int col, char val, vector<vector<char>>& board){// 判断行同一行是否有重复元素for(int i = 0; i < 9; i++){if(board[row][i] == val) return false;}// 判断行同一列是否有重复元素for(int j = 0; j < 9; j++){if(board[j][col] == val) return false;}// 判断所在9宫格内是否有重复元素(注意确定行列起点)int startRow = (row / 3) * 3;int startCol = (col / 3) * 3;for(int i = startRow; i < startRow + 3; i++){for(int j = startCol; j < startCol + 3; j++){if(board[i][j] == val) return false;}}return true;}
};