LeetCode-37. Sudoku Solver

0.原题

Write a program to solve a Sudoku puzzle by filling the empty cells.

A sudoku solution must satisfy all of the following rules:

1. Each of the digits 1-9 must occur exactly once in each row.
2. Each of the digits 1-9 must occur exactly once in each column.
3. Each of the the digits 1-9 must occur exactly once in each of the 9 3x3 sub-boxes of the grid.

Empty cells are indicated by the character '.'.

A sudoku puzzle...

...and its solution numbers marked in red.

Note:

• The given board contain only digits 1-9 and the character '.'.
• You may assume that the given Sudoku puzzle will have a single unique solution.
• The given board size is always 9x9.

 

1.代码

class Solution:
    def solveSudoku(self, board):
        """
        :type board: List[List[str]]
        :rtype: void Do not return anything, modify board in-place instead.
        """
        self.row = [[] for _ in range(9)]
        self.column = [[] for _ in range(9)]
        self.sub_box = [[[] for _ in range(3)]for _ in range(3)]
        for i in range(9):
            for j in range(9):
                if board[i][j] != '.':
                    self.row[i].append(board[i][j])
                    self.column[j].append(board[i][j])
                    sub_box_row = i // 3
                    sub_box_column = j // 3
                    self.sub_box[sub_box_row][sub_box_column].append(board[i][j])
        self.fun(board,0,0)

    def fun(self,board,row,column):
        if row == 9:
            return True
        else:
            if board[row][column] == '.':
                for i in ['1','2','3','4','5','6','7','8','9']:
                    sub_box_row = row // 3
                    sub_box_column = column // 3
                    if (i not in self.row[row]) and (i not in self.column[column]) and (i not in self.sub_box[sub_box_row][sub_box_column]):
                        board[row][column] = i
                        self.row[row].append(i)
                        self.column[column].append(i)
                        self.sub_box[sub_box_row][sub_box_column].append(i)
                        finish = self.next_square(board, row, column)
                        if finish:
                            return True
                        self.sub_box[sub_box_row][sub_box_column].pop()
                        self.column[column].pop()
                        self.row[row].pop()
                        board[row][column] = '.'
                return False
            else:
                return(self.next_square(board,row,column))

    def next_square(self,board,row=0,column=0):
        if column == 8:
            return(self.fun(board, row + 1, 0))
        else:
            return(self.fun(board, row, column + 1))

一开始没注意题目要求：rtype: void Do not return anything, modify board in-place instead.

结果查了很久的错，晕！

本题的思路与八皇后问题类似，仍然是使用递归遍历的思量，使用辅助数组进行剪枝操作。

八皇后问题请移步:https://blog.youkuaiyun.com/qq_17753903/article/details/82792082