leetcodeday37 –解数独

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：

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：回溯法：通俗理解就是如果board[i][j]=VALUE不满足条件就回退到上一步的选择，重新选择。

回溯法（探索与回溯法）是一种选优搜索法，又称为试探法，按选优条件向前搜索，以达到目标。但当探索到某一步时，发现原先选择并不优或达不到目标，就退回一步重新选择，这种走不通就退回再走的技术为回溯法，而满足回溯条件的某个状态的点称为“回溯点”。

代码实现：（递归）【参考了解题思路】

# [37] 解数独
#
#回溯法
"""
回溯法（探索与回溯法）是一种选优搜索法，又称为试探法，
按选优条件向前搜索，以达到目标。但当探索到某一步时，
发现原先选择并不优或达不到目标，就退回一步重新选择，
这种走不通就退回再走的技术为回溯法，
而满足回溯条件的某个状态的点称为“回溯点”。

"""
# @lc code=start
class Solution:
    def solveSudoku(self, board)->None:
        """
        Do not return anything, modify board in-place instead.
        """
        #判断改行、列、3*3小格子是否满足数独规则：

        def isRowSafe(row,value):
            for i in range(9):
                if board[row][i]==value:
                    return False
            return True
    
        def isColSafe(col,value):
            for i in range(9):
                if board[i][col]==value:
                    return False
            return True
        
        def isSmallboxSafe(row,col,value):
            inirow=row//3*3
            inicol=col//3*3
            for i in range(3):
                for j in range(3):
                    if board[i+inirow][j+inicol]==value:
                        return False  
            return True
        #判断该位置是否可行
        def isSafe(row,col,value):
            return isRowSafe(row,value) and isColSafe(col,value) and isSmallboxSafe(row,col,value)                 
        #解数独，结束条件
        def solve(row,col):
            if row==8 and col ==9:
                return True
            if col ==9:
                col=0
                row+=1
            if board[row][col]!=".":
                return solve(row,col+1)
            for i in range(1,10):
                if isSafe(row,col,str(i)):
                    i=str(i)
                    board[row][col] = i
                    if solve(row, col+1):
                        return board
           #回溯到上一个状态（也就是前一个solve）
            board[row][col]="."
            return False
        solve(0,0)
        print(board)