为什么迷宫求解器代码不起作用

Question

很抱歉格式错误和大量代码，我只是一个初学者，不知道如何将错误诊断为少量代码。我哪里出错了，它只是返回"在 0， 0 处发现为空"并退出。我已经查看了这种性质下的其他代码，无法理解给出的答案，需要对此特定代码进行更多阐述。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

没有声明这是否比其他答案更快，但是如果您想在找到出口时返回 True/False，那么您可以为每个递归调用维护一个布尔值。

（此外，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

Answer 2

return

是函数的终点，在代码中，您需要删除 return true，因为它会导致程序过早退出。另外，您需要一个 if 大小写来处理已经访问过的单元格（设置为 'x'）。

我玩了一分钟，最终结果是这样的：

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，即使在找到末尾后也会继续。如果你想要一些东西来实际解决你的迷宫并给你必要的步骤，你可以对此代码进行一些修改，以便它存储你采取的路径。您还可以研究广度优先搜索，因为这几乎是您在这里使用的。