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#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)] = a15 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)] = b21 22         return a + b 23 24 foo = Solution()25 print foo.uniquePaths(1, 2)