python - 将多个返回语句拆分为不同的函数

Question

我正在实现回溯解决方案并有多个返回语句。我没有看到将其拆分为多个函数的方法，因此每个函数只有一个返回语句。代码是..

  def solve_grid(self, grid, row=0, col=0):


    row, col = self.find_next(grid, row, col)
    if row == -1:
        return True
    for num in range(1,10):
        if self.isValid(grid, row, col, num):
            grid[row][col] = num
            if self.solve_grid(grid, row, col):
                return True
            grid[row][col] = 0
    return False

我试过按如下方式拆分它

def check(self, grid, row, col):
    boolean = None
    row, col = self.find_next(grid, row, col)
    if row == -1:
        boolean = True
    return boolean

def solve_grid(self, grid, row=0, col=0):

    boolean = None
    if not self.check(grid, row, col):
        for num in range(1,10):
            if self.isValid(grid, row, col, num):
                grid[row][col] = num
                if self.solve_grid(grid, row, col):
                    boolean = True
                else: 
                    boolean = False
            grid[row][col] = 0
    return boolean

这会导致最大递归深度。我对如何解决这个问题有点迷茫，我以前从来没有真正尝试过拆分多个 return 语句。任何指示或提示都会有所帮助。

Answer 1

如果您想做的只是删除多个返回值，这就可以了

def solve_grid(self, grid, row=0, col=0):
    row, col = self.find_next(grid, row, col)
    if row == -1:
        result = True
    else:
        result = False
        for num in range(1,10):
            if self.isValid(grid, row, col, num):
                grid[row][col] = num

                if self.solve_grid(grid, row, col):
                    result=True
                    break

                grid[row][col] = 0

    return result

您还可以将 for 循环转换为 while 以移除 break

def solve_grid(self, grid, row=0, col=0):
    row, col = self.find_next(grid, row, col)
    if row == -1:
        result = True
    else:
        result = False
        num = 0
        while num < 9 and not result:
            num += 1
            if self.isValid(grid, row, col, num):
                grid[row][col] = num

                if self.solve_grid(grid, row, col):
                    result=True
                else:
                    grid[row][col] = 0

    return result

但我个人认为您的原始表格更具可读性。

最后的简化是通过检查 row 来初始化 result，从而消除了一定程度的缩进

def solve_grid(self, grid, row=0, col=0):
    row, col = self.find_next(grid, row, col)
    result = (row == -1)
    num = 0
    while num < 9 and not result:
        num += 1
        if self.isValid(grid, row, col, num):
            grid[row][col] = num

            if self.solve_grid(grid, row, col):
                result=True
            else:
                grid[row][col] = 0


    return result

现在，我觉得还算干净