题解 | #不同路径的数目(一)#
不同路径的数目(一)
https://www.nowcoder.com/practice/166eaff8439d4cd898e3ba933fbc6358
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]
