问题描述
我已经在python中实现了回溯算法来解决数独。在算法的步骤中,我检查数独是否已经解决,何时解决,我想将其返回。但是我的函数只是返回None。有人可以解释一下原因,并给我提示解决方案吗?
#Test-Sudoku
grid = [[1,7,5,4,0],[9,3,[0,2,8,9,6,1,1],7]]
def possible(row,col,num,grid):
""" Check if num can be passed to grid[row][col]"""
if grid[row][col] != 0:
return False
for i in range(9):
if grid[row][i] == num or grid[i][col] == num:
return False
row0 = (row//3)*3
col0 = (col//3)*3
for i in range(3):
for j in range(3):
if grid[row0+i][col0+j] == num:
return False
return True
def is_solved(grid):
""" Check if Sudoku is already solved/full"""
for i in range(9):
for j in range(9):
if grid[i][j] == 0:
return False
return True
def solve(grid):
""" Backtracking algorithm to solve Sudoku"""
for r in range(9):
for c in range(9):
if grid[r][c] == 0:
for i in range(1,10):
if possible(r,c,i,grid):
grid[r][c] = i
if is_solved(grid):
# Print grid for test
print("Show Solved Sudoku:")
print(grid)
return(grid)
solve(grid)
grid[r][c] = 0
return
return
solved_sudoku = solve(grid)
print(solved_sudoku)
解决方法
因为您使用return
返回none
您将结果存储在grid
中,而grid
的变化每转
尝试一下:
def solve(grid):
""" Backtracking algorithm to solve Sudoku"""
for r in range(9):
for c in range(9):
if grid[r][c] == 0:
for i in range(1,10):
if possible(r,c,i,grid):
grid[r][c] = i
solve(grid)
if is_solved(grid):
return
grid[r][c] = 0
return
solve(grid)
print(grid)