Python-迷宫问题-使用栈解决迷宫问题

 

# 迷宫
maze = [
    [1, 1, 1, 1, 1, 1, 1, 1, 1, 1],
    [1, 0, 1, 1, 1, 1, 0, 1, 1, 1],
    [1, 0, 1, 1, 1, 1, 0, 1, 0, 1],
    [1, 0, 1, 0, 0, 0, 0, 0, 0, 1],
    [1, 0, 0, 0, 1, 0, 1, 1, 1, 1],
    [1, 0, 1, 0, 1, 0, 1, 1, 0, 1],
    [1, 1, 1, 1, 1, 0, 0, 0, 0, 1],
    [1, 1, 1, 1, 1, 1, 1, 1, 1, 1],
]

# 封装四个方向
dirs = [
    lambda x, y: (x + 1, y),
    lambda x, y: (x - 1, y),
    lambda x, y: (x, y - 1),
    lambda x, y: (x, y + 1)
]


def maze_path(x1, y1, x2, y2):
    """
    :param x1: 起点横坐标
    :param y1: 起点纵坐标
    :param x2: 终点横坐标
    :param y2: 终点纵坐标
    :return:
    """
    global nextNode
    stack = []
    stack.append((x1, y1))
    while (len(stack) > 0):
        curNode = stack[-1]  # 当前节点
        if curNode[0] == x2 and curNode[1] == y2:
            # 走到终点了
            for p in stack:
                print(p)
            return True

        for dir in dirs:
            nextNode = dir(curNode[0], curNode[1])
            # 如果下一个节点能走
            if maze[nextNode[0]][nextNode[1]] == 0:
                stack.append(nextNode)
                maze[nextNode[0]][nextNode[1]] = 2  # 2表示为已经走过的地方
                break
        else:
            maze[nextNode[0]][nextNode[1]] = 2
            stack.pop()  # 在找不到路走的情况下，进行回退上一节点

    else:
        print("没路")
        return False


maze_path(1, 1, 6, 8)