题解 | 不同路径的数目(一)

#
# 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
#
# 
# @param m int整型 
# @param n int整型 
# @return int整型
#
class Solution:
    def uniquePaths(self , m: int, n: int) -> int:
        # write code here
        dp = [[0 for _ in range(n+1)] for _ in range(m+1)]
       
        for i in range(1,n+1):
            dp[1][i] = 1
        for j in range(1,m+1):
            dp[j][1] = 1
        for i in range(2, m+1):
            for j in range(2, n+1):
                dp[i][j] = dp[i-1][j] + dp[i][j-1]
        return dp[m][n]