Lintcode 611. Knight Shortest Pa

求knight从source 到destination的最短路径。采用bfs和dictionary，dictionary用来记录点到（x，y）的最短路径。
算法：首先把source放入queue中，并记录source对应路径为0，将source周围一步可以到达的点放入queue中，并记录他们的路径长度为1，继续上述操作，如果点超出矩阵或者矩阵对应点为1（barriar）或者已经走过此点，则跳过上述操作。最后如果queue中弹出的点是destination，则对应路径为最短路径。如果没有返回最短路径则说明到不了这点，则返回-1。
python代码：

class Solution:
    """
    @param grid: a chessboard included 0 (false) and 1 (true)
    @param source: a point
    @param destination: a point
    @return: the shortest path 
    """
    def shortestPath(self, grid, source, destination):
        # write your code here
        from collections import deque
        delta = [[1, 2], [1, -2], [-1, 2], [-1, -2], [2, 1], [2, -1], [-2, 1], [-2, -1]]
        dictionary = {(source.x, source.y): 0}
        queue = deque()
        queue.append([source.x, source.y])
        while queue:
            x, y = queue.popleft()
            if (x, y) == (destination.x, destination.y):
                return dictionary[(x, y)]
            for delta_x, delta_y in delta:
                newx = x + delta_x
                newy = y + delta_y
                if newx < 0 or newx >= len(grid) or newy < 0 or newy >= len(grid[0]):
                    continue
                if grid[newx][newy] == 1:
                    continue
                if (newx, newy) in dictionary:
                    continue
                queue.append([newx, newy])
                dictionary[(newx, newy)] = dictionary[(x, y)] + 1
        return -1