为什么不会迷宫解算器代码工作

Question

抱歉格式错误和代码量大，我只是初学者，不知道如何诊断错误到更小的代码量。我哪里出错了它只是 returns "found empty at 0, 0" 然后退出。我已经查看过这种性质的其他代码，无法理解给出的答案，需要对此特定代码进行更多详细说明。请帮忙

 grid = [["T", " ", " ", "#", "#", "#"],
        ["#", "#", " ", " ", " ", "#"],
        ["#", " ", " ", "#", " ", "#"],
        ["#", "#", "#", " ", " ", "#"],
        ["#", "#", " ", " ", "#", " "],
        ["#", "#", "#", "#", "#", "E"]]

def print_grid():
    pr_grid = ""
    for key in grid:
        for num in key:
            pr_grid += str(num)
            pr_grid += " "
        print pr_grid
        pr_grid = ""

    print pr_grid

run = True

def main(x, y):
    if grid[x][y] == " " or grid[x][y] == "T":
        print "Found empty at %d %d" % (x, y)
        grid[x][y] = "x"
    elif grid[x][y] == "#":
        print "Found wall at %d %d" % (x, y)
    elif grid[x][y] == "E":
        print "Found exit at %d %d" % (x, y)

    if y < len(grid)-1:
        main(x, y + 1)
    if y > 0:
        main(x, y - 1)
    if x < len(grid[x])-1:
        main(x + 1, y)
    if x > 0:
        main(x - 1, y)


print_grid()

main(0, 0)

print_grid()

Answer 1

return 是函数的终点，在您的代码中您需要删除 return true 因为它会导致您的程序过早退出。此外，您需要一个 if 案例来处理已经访问过的单元格（设置为 'x'）。

我玩了一分钟，最后的结果是这样的：

def main(x, y):
    if grid[x][y] == " " or grid[x][y] == "T":
        print "Found empty at %d %d" % (x, y)
        grid[x][y] = "x"
    elif grid[x][y] == "#":
        print "Found wall at %d %d" % (x, y)
        return
    elif grid[x][y] == "E":
        print "Found exit at %d %d" % (x, y)
        return 
    else: return

    if y < len(grid)-1:
        main(x, y + 1)
    if y > 0:
        main(x, y - 1)
    if x < len(grid[x])-1:
        main(x + 1, y)
    if x > 0:
        main(x - 1, y)

请注意，此代码查找并访问网格中的每个单元格，并在找到末尾后继续执行。如果你想要一些东西来真正解决你的迷宫并给你必要的步骤，你可以对这段代码做一些修改，以便它存储你所走的路径。您还可以研究 Breadth-First-Search，因为这几乎就是您在这里使用的内容。

Answer 2

没有关于这是否比其他答案更快的说法，但是如果你想 return True/False 如果你找到了出口，那么你可以为每个递归调用。

（另外，您的 print_grid 方法可以更短）

def print_grid():
    print "\n".join(' '.join(row) for row in grid)

无论如何，这是我对程序所做的修改。

def main(x, y):
    # Check going out of bounds
    if y < 0 or y >= len(grid): 
        return False
    if x < 0 or x >= len(grid[y]):
        return False

    if grid[x][y] == "E":
        print "Found exit at %d %d" % (x, y)
        return True
    elif grid[x][y] == "#":
        print "Found wall at %d %d" % (x, y)
        return False
    elif grid[x][y] == " " or grid[x][y] == "T":
        print "Found empty at %d %d" % (x, y)
        grid[x][y] = "x"
        # no return, we want to continue searching
    else: # catch invalid characters
        return False

    found = False
    # "Bubble-up" the results from searching for the exit 
    # Also limit the search space by keeping track if the exit was found
    if y < len(grid)-1 and not found:
        found = main(x, y + 1)
    if y > 0 and not found:
        found = main(x, y - 1)
    if x < len(grid[x])-1 and not found:
        found = main(x + 1, y)
    if x > 0 and not found:
        found = main(x - 1, y)

    return found

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