37. 解数独

Difficulty

Hard

Tags

回溯

URL

https://leetcode-cn.com/problems/sudoku-solver/

Star

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：


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

通过次数97,273提交次数145,034

回溯

思路

其实就是把选择路径范围扩大到二维了，其他的没什么区别

题解


class Solution:
    def solveSudoku(self, board: List[List[str]]) -> None:
        """
        Do not return anything, modify board in-place instead.
        """
        def backtrack(board):
            for i in range(len(board)):  #遍历行
                for j in range(len(board[0])):  #遍历列
                    if board[i][j] != ".": continue
                    for k in range(1,10):  #(i, j) 这个位置放k是否合适
                        if isValid(i,j,k,board):
                            board[i][j] = str(k)  #放置k
                            if backtrack(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
        def isValid(row,col,val,board):
            for i in range(9):  #判断行里是否重复
                if board[row][i] == str(val):
                    return False
            for j in range(9):  #判断列里是否重复
                if board[j][col] == str(val):
                    return False
            startRow = (row // 3) * 3
            startcol = (col // 3) * 3
            for i in range(startRow,startRow + 3):  #判断9方格里是否重复
                for j in range(startcol,startcol + 3):
                    if board[i][j] == str(val):
                        return False
            return True
        backtrack(board)

可以不用专门封装这个函数的，如下：


class Solution:
    def solveSudoku(self, board: List[List[str]]) -> None:
        """
        Do not return anything, modify board in-place instead.
        """
        for i in range(len(board)):  
            for j in range(len(board[0])):  
                if board[i][j] == ".": 
                    for k in range(1,10):  
                        if self.isValid(i,j,k,board):
                            board[i][j] = str(k) 
                            if self.solveSudoku(board): 
                                return True  
                            board[i][j] = "."  
                    return False  
        return True  
        
    def isValid(self, row,col,val,board):
        for i in range(9):  #判断行里是否重复
            if board[row][i] == str(val):
                return False
        for j in range(9):  #判断列里是否重复
            if board[j][col] == str(val):
                return False

        startRow = (row // 3) * 3
        startcol = (col // 3) * 3
        for i in range(startRow,startRow + 3):  #判断9方格里是否重复
            for j in range(startcol,startcol + 3):
                if board[i][j] == str(val):
                    return False
        return True