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:
        dp = [[0 for _ in range(n)] for _ in range(m)]#创建dp数组，初始化数值为0
        dp[0][0] = 1#初始化
        for j in range(1, n):
            dp[0][j] = dp[0][j-1]  # 只能从左边来
        for i in range(1, m):
            dp[i][0] = dp[i-1][0]  # 只能从上边来
        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[m-1][n-1]