LeetCode668马在棋盘上的概率

class Solution(object):
    def knightProbability(self, N, K, r, c):
        """
        :type N: int
        :type K: int
        :type r: int
        :type c: int
        :rtype: float
        """
        dp = []
        count = 0.0
        for i in range(K+1):
            temp1 = []
            for j in range(N):
                temp1.append([0] * N)
            dp.append(temp1)
        dp[0][r][c] = 1
        directions = [(1,2),(1,-2),(2,1),(2,-1),(-1,2),(-1,-2),(-2,1),(-2,-1)]
        for n in range(1,K+1):
            for x in range(N):
                for y in range(N):
                    for (X,Y) in directions:
                        if (x+X) >= 0 and (x+X) < N and (y+Y) >= 0 and (y+Y) < N:
                            dp[n][x][y] += dp[n-1][x+X][y+Y]                      #计算第n次棋盘中每个位置可能出现的次数
        for i in dp[-1]:
            for j in i:
                count += j
        return count / 8 ** K