考虑到起始位置是r，c，我计算在K步之后停留N个N棋盘的概率。这是我的代码：
def knightProbability(self, N, K, r, c):
# dfs.
answer = 0
dp = [[[-1] * N]*N] * (K+1)
# dfs with memoization.
self.dic = {}
return self.dfs(N, K, r, c, dp) # Starting at r, c, what's the probability of staying on board after k moves.
def dfs(self, N, k, r, c, dp):
offsets = [(-1, -2), (-1, 2), (-2, -1), (-2, 1), (1, -2), (1, 2), (2, -1), (2, 1)]
answer = 0.0
if k == 0:
return 1.0
'''if dp[k][r][c] >= 0:
return dp[k][r][c]'''
if str(k) + ":" + str(r) + ":" + str(c) in self.dic:
return self.dic[str(k) + ":" + str(r) + ":" + str(c)]
for offset in offsets:
new_r = r + offset[0]
new_c = c + offset[1]
if new_r > -1 and new_r < N and new_c > -1 and new_c < N:
answer += self.dfs(N, k-1, new_r, new_c, dp) * 1.0 / 8 # with k-1 steps, what's the probability?
'''dp[k][r][c] = answer'''
self.dic[str(k) + ":" + str(r) + ":" + str(c)] = answer
return answer
这是一个简单的dfs。当我使用字典来存储和检索我现在正在做的访问状态（k，r，c）时，它可以工作。但是，如果我使用k by N by N数组来存储和检索它们，我得不到正确的答案。有什么想法吗？注释部分是我正在讨论的数组部分的存储/检索，它应该与当前代码做同样的事情。