题目
37. 解数独
困难
相关标签
数组 哈希表 回溯 矩阵
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
思路和解题方法
- 遍历行和列,找到一个空白格子(
'.'
)。- 从 1 到 9 依次尝试填入数字,判断当前数字是否合适放置在该格子位置。
- 在
isValid
函数中,首先检查当前数字在所在行和列是否重复出现,如果重复则不合适。然后确定当前格子所在的 3x3 小方格起始位置,再检查该小方格内是否存在重复的数字。- 如果当前数字合适,则将其填入格子中,并继续递归地调用回溯函数
backtracking
,以尝试解决剩下的空白格子。- 如果递归函数返回 true,说明成功找到了一组合适的数独解法,直接返回即可。
- 如果递归函数返回 false,说明当前尝试的数字不合适,需要进行回溯操作,撤销当前数字的填入,并继续尝试下一个数字。
- 如果遍历完所有的格子都没有返回 true,说明无法得到有效解,整个算法结束。
最终,当算法完成后,数独棋盘中的空白格子将被填满,或者数独问题没有解。
c++ 代码
class Solution {
private:
bool backtracking(vector<vector<char>>& board) {
for (int i = 0; i < board.size(); i++) { // 遍历行
for (int j = 0; j < board[0].size(); j++) { // 遍历列
if (board[i][j] == '.') {
for (char k = '1'; k <= '9'; k++) { // (i, j) 这个位置放k是否合适
if (isValid(i, j, k, board)) {
board[i][j] = k; // 放置k
if (backtracking(board)) return true; // 如果找到合适一组立刻返回
board[i][j] = '.'; // 回溯,撤销k
}
}
return false; // 9个数都试完了,都不行,那么就返回false
}
}
}
return true; // 遍历完没有返回false,说明找到了合适棋盘位置了
}
bool isValid(int row, int col, char val, vector<vector<char>>& board) {
for (int i = 0; i < 9; i++) { // 判断行里是否重复
if (board[row][i] == val) {
return false;
}
}
for (int j = 0; j < 9; j++) { // 判断列里是否重复
if (board[j][col] == val) {
return false;
}
}
int startRow = (row / 3) * 3;
int startCol = (col / 3) * 3;
for (int i = startRow; i < startRow + 3; i++) { // 判断9方格里是否重复
for (int j = startCol; j < startCol + 3; j++) {
if (board[i][j] == val ) {
return false;
}
}
}
return true;
}
public:
void solveSudoku(vector<vector<char>>& board) {
backtracking(board);
}
};
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