leetcode题解62：不同路径

描述

分析

最经典也最基础的动态规划题了，用dp[i][j]表示到达i,j这个网格的步数，这有转移方程

dp[i][j] = \left\{ \begin{aligned} & 1 & i == 0\text{ or }j == 0 \\ & dp[i-1][j] + dp[i][j-1] & others \end{aligned}\right.

即步数总是等于到达相邻的上方与左方的步数之和。

代码

python

class Solution:
    def uniquePaths(self, m: int, n: int) -> int:
        dp = [[0 for _ in range(n)] for _ in range(m)]
        for i in range(m):
            for j in range(n):
                if i == 0 or j == 0:
                    dp[i][j] = 1
                else:
                    dp[i][j] = dp[i-1][j] + dp[i][j-1]
        return dp[m-1][n-1]