#Leet Code# Unique Path

描述：

使用了递归，有些计算是重复的，用了额外的空间，Version 1是m*n

Bonus：一共走了m+n步，例如 m = 2, n = 3 [#, @, @, #, @]，所以抽象成数学问题，解是C（m + n, m）

代码：

 1 class Solution:
 2     # @return an integer
 3     def __init__(self):
 4         self.record = {}
 5
 6     def uniquePaths(self, m, n):
 7         if m == 0 or n == 0: return 0
 8         if m == 1 or n == 1: return 1
 9
10         if (m-1, n) in self.record:
11             a = self.record[(m-1, n)]
12         else:
13             a = self.uniquePaths(m-1, n)
14             self.record[(m-1, n)] = a
15
16         if (m, n-1) in self.record:
17             b = self.record[(m, n-1)]
18         else:
19             b = self.uniquePaths(m, n-1)
20             self.record[(m, n-1)] = b
21
22         return a + b
23
24 foo = Solution()
25 print foo.uniquePaths(1, 2)

#Leet Code# Unique Path