A robot is located at the top-left corner of a m x n grid (marked 'Start' in the diagram below).
The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid (marked 'Finish' in the diagram below).
How many possible unique paths are there?
Above is a 7 x 3 grid. How many possible unique paths are there?
Note: m and n will be at most 100.
Example 1:
Input: m = 3, n = 2
Output: 3
Explanation:
From the top-left corner, there are a total of 3 ways to reach the bottom-right corner:
1. Right -> Right -> Down
2. Right -> Down -> Right
3. Down -> Right -> Right
Example 2:
Input: m = 7, n = 3
Output: 28
题意
机器人走m * n的格子，只能向右或者向下，从左上角走到右下角，共有多少种走法。
思路
动态规划，最右下角只能从它的上面或者左边到达，所以状态方程为：
dp[i][j] = dp[i-1][j] + dp[i][j-1]
边界：m为1， 或者 n 为1时，只有一种情况。
代码
class Solution:
def uniquePaths(self, m: int, n: int) -> int:
if m == 1 or n == 1:
return 1
dp = [[1] * n for _ in range(m)]
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]