leetcode-37.解数独
深度优先搜索(dfs)
回溯法
题目详情
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:
1.数字 1-9
在每一行只能出现一次。
2.数字 1-9
在每一列只能出现一次。
3.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:
我的代码:
class Solution
{
public:
bool backtracking(vector<vector<char>>& board)
{ //遍历
for(int i = 0; i < board.size(); ++i)
{
for(int j = 0; j < board[0].size(); ++j)
{
if(board[i][j] != '.') continue;
for(char k = '1'; k <= '9'; ++k) //在[i,j]处依次尝试填入1-9
{
if(isValid(i, j, k, board)) //如果填入后满足数独条件
{
board[i][j] = k; //那就填入
if(backtracking(board)) return true; //然后递归接着填,如果都填完就会一直true回来
board[i][j] = '.'; //回溯,撤销填入的k
}
}
return false; //如果能运行到这里,那么肯定上面1-9没有一个满足条件
}
}
return true; //如果能运行到这里,那么上面没有false肯定有答案
}
bool isValid(int row, int col, char val, vector<vector<char>>& board)
{
for(int i = 0; i < 9; ++i) //1. 行必须无重复
{
if(board[row][i] == val)
return false;
}
for(int j = 0; j < 9; ++j) //2. 列必须无重复
{
if(board[j][col] == val)
return false;
}
//3. 3*3格内必须无重复
//startRow startCol计算出[row,col]所在九宫格的初始行列点
int stratRow = (row / 3) * 3;
int startCol = (col / 3) * 3;
for(int i = stratRow; i < stratRow + 3; ++i)
{
for(int j = startCol; j < startCol + 3; ++j)
{
if(board[i][j] == val)
return false;
}
}
return true;
}
void solveSudoku(vector<vector<char>>& board)
{
backtracking(board);
}
};
涉及知识点:
1.深度优先搜索(dfs)
深度优先搜索(depth-first seach,DFS)在搜索到一个新的节点时,立即对该新节点进行遍历;因此遍历需要用先入后出的栈来实现,也可以通过与栈等价的递归来实现。对于树结构而言,由于总是对新节点调用遍历,因此看起来是向着“深”的方向前进。