# \[62]\[中等]\[动态规划] 不同路径

## 题目描述

[62. 不同路径](https://leetcode-cn.com/problems/unique-paths/)

一个机器人位于一个 m x n 网格的左上角 （起始点在下图中标记为“Start” ）。

机器人每次只能向下或者向右移动一步。机器人试图达到网格的右下角（在下图中标记为“Finish”）。

问总共有多少条不同的路径？

![](/files/-MI7NvNKuIiaukKrqfz0)

例如，上图是一个7 x 3 的网格。有多少可能的路径？

示例 1:

```
输入: m = 3, n = 2
输出: 3
解释:
从左上角开始，总共有 3 条路径可以到达右下角。
1. 向右 -> 向右 -> 向下
2. 向右 -> 向下 -> 向右
3. 向下 -> 向右 -> 向右
```

示例 2:

```
输入: m = 7, n = 3
输出: 28
```

提示：

* 1 <= m, n <= 100
* 题目数据保证答案小于等于 2 \* 10 ^ 9

## 解题思路

### 动态规划

```python
class Solution:
    def uniquePaths(self, m: int, n: int) -> int:
        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[-1][-1]
```

### 滚动数组

可以看到状态转移公式只是用`dp[i - 1][j]`和`dp[i][j - 1]`的值, 因此用一列的长度, 配合从上到下循环(或一行的长度, 从左到右循环)就足够了. 对应的空间复杂度为$$O(\min({m,n}))$$

```python
class Solution:
    def uniquePaths(self, m: int, n: int) -> int:
        if m > n:
            m, n = n, m
        dp = [1] * m

        for _ in range(1, n):
            for i in range(1, m):
                dp[i] = dp[i] + dp[i - 1]
        return dp[-1]
```


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