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
        '''
        if m==1 or n==1:
            return 1
        else:
            return self.uniquePaths(m-1,n) +self.uniquePaths(m,n-1)
        '''
        dp = [[0] * (n + 1) for i in range(m + 1)]
        print(dp)
        for i in range(1, m+1):
            for j in range(1,n+1):
                if i == 1:
                    dp[i][j] =1
                    continue
                if j == 1:
                    dp[i][j] = 1
                    continue
                dp[i][j] = dp[i-1][j] + dp[i][j-1]
        return dp[m][n]