The Maze

490. The Maze

Given the ball's start position, the destination and the maze, determine whether the ball could stop at the destination.

"""
利用while loop到達牆壁
只紀錄端點的情況
"""
def hasPath(self, maze: List[List[int]], start: List[int], destination: List[int]) -> bool:
    m, n = len(maze), len(maze[0])
    visited = set(tuple(start))
    q = [start]
    
    while q:
        pos = q.pop(0)
        
        for dy, dx in ((0, 1), (1, 0), (0, -1), (-1, 0)):
            y = pos[0] + dy
            x = pos[1] + dx
            inloop = False
            
            while 0 <= y < m and 0 <= x < n and maze[y][x] == 0:
                inloop = True
                if pos == destination: return True
                y += dy
                x += dx
            
            # 記得退一步
            end = [y-dy, x-dx]
            if inloop and tuple(end) not in visited:
                q.append(end)
                visited.add(tuple(end))
                
    return False

505. The Maze II

Given the ball's start position, the destination and the maze, find the shortest distance for the ball to stop at the destination. The distance is defined by the number of empty spaces traveled by the ball from the start position (excluded) to the destination (included). If the ball cannot stop at the destination, return -1.

def shortestDistance(self, maze: List[List[int]], start: List[int], destination: List[int]) -> int:
    m, n = len(maze), len(maze[0])
    dist = [[float('inf')] * n for _ in range(m)]
    dist[start[0]][start[1]] = 0
    q = [(start[0], start[1], 0)]
    d = destination
    
    while q:
        p1, p2, step = q.pop(0)
        for dy, dx in ((0, 1), (1, 0), (0, -1), (-1, 0)):
            s = step + 1
            y = p1 + dy
            x = p2 + dx
            
            found = False
            inloop = False
            
            while 0 <= y < m and 0 <= x < n and maze[y][x] == 0:
                inloop = True
                if (y, x) == d and s < dist[y][x]:
                    dist[y][x] = [y, x]
                    found = True
                    break       
                s += 1
                y += dy
                x += dx
                
            # 記得退一步
            y, x = y-dy, x-dx
            s -= 1

            if inloop and not found and s < dist[y][x]:
                dist[y][x] = s
                q.append((y, x, s))

    return dist[d[0]][d[1]] if dist[d[0]][d[1]] != float('inf') else -1

499. The Maze III

Given the ball position, the hole position and the maze, find out how the ball could drop into the hole by moving the shortest distance. The distance is defined by the number of empty spaces traveled by the ball from the start position (excluded) to the hole (included). Output the moving directions by using 'u', 'd', 'l' and 'r'. Since there could be several different shortest ways, you should output the lexicographically smallest way. If the ball cannot reach the hole, output "impossible".

 def findShortestWay(self, maze: List[List[int]], ball: List[int], hole: List[int]) -> str:
    m, n = len(maze), len(maze[0])
    dist = [[(float('inf'), [])] * n for _ in range(m)]
    q = [(ball[0], ball[1], 0, [''])]
    dirs = {0:'r', 1:'d', 2:'l', 3:'u'}
    
    while q:
        p1, p2, step, routes = q.pop(0)
        for i, (dy, dx) in enumerate(((0, 1), (1, 0), (0, -1), (-1, 0))):
            s = step+1
            y = p1+dy
            x = p2+dx
            new_routes = [r+dirs[i] for r in routes]
            found = False
            inloop = False
            while 0 <= y < m and 0 <= x < n and maze[y][x] == 0:
                inloop = True
                if [y, x] == hole:
                    if s == dist[y][x][0]:
                        dist[y][x] = (s, dist[y][x][1] + new_routes)
                    elif s < dist[y][x][0]:
                        dist[y][x] = (s, new_routes)
                    found = True
                    break
                s += 1
                y += dy
                x += dx
                     
            # 記得退一步
            y, x = y-dy, x-dx
            s -= 1
            if inloop and not found and s <= dist[y][x][0]:
                if s == dist[y][x][0]:
                    dist[y][x] = (s, dist[y][x][1] + new_routes)
                elif s < dist[y][x][0]:
                    dist[y][x] = (s, new_routes)
                q.append((y, x, s, new_routes))

    if dist[hole[0]][hole[1]][0] == float('inf'): return 'impossible'
    return sorted(dist[hole[0]][hole[1]][1])[0]