leetcode62 不同路徑

要走到最下角，一共要走m-1+n-1=m+n-2步，即從m+n-2中挑出m-1步向下走就可以了，或者挑出n-1步向右走（排列組合式子等值）。那麼按照這樣，所有的走法即為

class solution:

def uniquepaths(self, m: int, n: int) -> int:
#定義計算排列組合的函式
def cheng(n):
res = 1
for i in range(n,0,-1):
res *= i
return res
#直接返回
return int(cheng(m+n-2) / (cheng(n-1) * cheng(m-1)))

class solution: 

def uniquepaths(self, m: int, n: int) -> int: 
if m == 0 or n == 0: 
return 0 
dp = [[0] * n for _ in range(m)]
#第一列
for i in range(m):
dp[i][0] = 1
#第一行
for j in range(n):
dp[0][j] = 1
for i in range(1,m):
for j in range(1,n):
dp[i][j] = dp[i-1][j] + dp[i][j-1]
return dp[-1][-1]