# 62.unique-paths

## Statement

• Difficulty: Medium
• Tag: `数学` `动态规划` `组合数学`

``````输入：m = 3, n = 7

``````输入：m = 3, n = 2

1. 向右 -> 向下 -> 向下
2. 向下 -> 向下 -> 向右
3. 向下 -> 向右 -> 向下
``````

``````输入：m = 7, n = 3

``````

``````输入：m = 3, n = 3

• `1 <= m, n <= 100`
• 题目数据保证答案小于等于 `2 * 109`

• Difficulty: Medium
• Tag: `Math` `Dynamic Programming` `Combinatorics`

There is a robot on an `m x n` grid. The robot is initially located at the top-left corner (i.e., `grid[0][0]`). The robot tries to move to the bottom-right corner (i.e., `grid[m - 1][n - 1]`). The robot can only move either down or right at any point in time.

Given the two integers `m` and `n`, return the number of possible unique paths that the robot can take to reach the bottom-right corner.

The test cases are generated so that the answer will be less than or equal to `2 * 109`.

Example 1:

``````Input: m = 3, n = 7
Output: 28
``````

Example 2:

``````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 -> Down -> Down
2. Down -> Down -> Right
3. Down -> Right -> Down
``````

Constraints:

• `1 <= m, n <= 100`

## Solution

``````class Solution:
def uniquePaths(self, m: int, n: int) -> int:
f = [[0 for i in range(n + 1)] for j in range(m + 1)]
f[0][0] = 1
for i in range(m):
for j in range(n):
if i - 1 >= 0:
f[i][j] += f[i - 1][j]
if j - 1 >= 0:
f[i][j] += f[i][j - 1]
return f[m - 1][n - 1]
``````
``````/**
* @param {string} s
* @param {string} p
* @return {boolean}
*/
var isMatch = function (s, p) {
const r = new RegExp(`^\${p}\$`);
return r.test(s);
};
``````