题解 | #不同路径的数目(一)#

class Solution:
    def uniquePaths(self, m: int, n: int) -> int:
        dp = [[0] * n for _ in range(m)]

        # 初始化第一行和第一列
        for i in range(m):
            dp[i][0] = 1
        for j in range(n):
            dp[0][j] = 1

        # 动态规划计算路径数量
        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]