m行n列, 需要往下走m-1步, 往右走n-1步, 也就是求C(m-1+n-1, n-1)或C(m-1+n-1, m-1)。
class Solution:
# @return an integer
def uniquePaths(self, m, n):
N = m - 1 + n - 1
K = min(m, n) - 1
# calculate C(N, K)
res = 1
for i in xrange(K):
res = res * (N - i) / (i + 1)
return res