哈尔123：```python # -*- coding:utf-8 -*- class Solution: def movingCount(self, k, r, c): # write code here self.cnt = 0 if r == 0 or c == 0: return 0 def p_sum(x, y): res = 0 while x: res += x%10 x = x//10 while y: res += y%10 y = y//10 return res <= k mat = [[1 for _ in range(c)] for _ in range(r)] def dfs(i, j): if not (0<=i<r and=""><=j</r>

lllong33：(死记)快慢指针。快指针一次走两步，慢指针一次走一步, 此时让快指针回到head, 继续走, 当两个指针再次相遇时恰好是在入口结点。

O_n2：关键在于递推公式的推导，这一篇讲得还蛮好 https://blog.csdn.net/crazy__chen/article/details/45115911

牛客5905718：这个用到了线性代数中的可逆，AB->BA (A^-1B^-1)^-1=BA; 其中A^-1:为A的逆