题解 | #Sudoku# 递归函数

l_r=[] # 每行 1*9
for i in range(9):
    s=list(map(int,input().split()))
    l_r.append(s)
#print(l_r)

def check(l_r,i,j):

    for k in range(9):
        if l_r[i][k]==l_r[i][j] and k!=j:
            return False
    for k in range(9):
        if l_r[k][j]==l_r[i][j] and k!=i:
            return False
    m = 3*(i // 3) #m,n分别是i,j位置所在的3*3格子的最左上角的位置
    n = 3*(j // 3)
    for k in range(3):
        for z in range(3):
            if (l_r[m+k][n+z] == l_r[i][j]) and ((m+k) != i) and ((n+z) != j): #判断所在3*3格子是否有重复的数据
                return False
    return True
 
def find_sudoku(l_r):

    num=[1,2,3,4,5,6,7,8,9]
    
    for i in range(9):
        for j in range(9):
            if l_r[i][j]==0:
                for n in num:
                    l_r[i][j]=n
                    if check(l_r,i,j) and find_sudoku(l_r):
                        #l_r[i][j]=n  #不能直接判定该位置可以填n
                        return True #n满足条件返回true
                    l_r[i][j]=0  #n不满足上述条件，取值恢复为0
                return False 
                # 所有n都不满足条件，返回false(该数独不合法)
    return True

find_sudoku(l_r)

for i in l_r:
    for j in i:
        print(j,end=' ')
    print()

'''
0 9 5 0 2 0 0 6 0
0 0 7 1 0 3 9 0 2
6 0 0 0 0 5 3 0 4
0 4 0 0 1 0 6 0 7
5 0 0 2 0 7 0 0 9
7 0 3 0 9 0 0 2 0
0 0 9 8 0 0 0 0 6
8 0 6 3 0 2 1 0 5
0 5 0 0 7 0 2 8 3
'''