solution.md

# 不同路径

<p>一个机器人位于一个 <code>m x n</code><em> </em>网格的左上角 （起始点在下图中标记为 “Start” ）。</p>
<p>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” ）。</p>
<p>问总共有多少条不同的路径？</p>
<p> </p>
<p><strong>示例 1：</strong></p><img
    src="https://cdn.jsdelivr.net/gh/doocs/leetcode@main/solution/0000-0099/0062.Unique%20Paths/images/robot_maze.png" />
<pre><strong>输入：</strong>m = 3, n = 7<strong><br />输出：</strong>28</pre>
<p><strong>示例 2：</strong></p>
<pre><strong>输入：</strong>m = 3, n = 2<strong><br />输出：</strong>3<strong><br />解释：</strong>从左上角开始，总共有 3 条路径可以到达右下角。<br />1. 向右 -> 向下 -> 向下<br />2. 向下 -> 向下 -> 向右<br />3. 向下 -> 向右 -> 向下</pre>
<p><strong>示例 3：</strong></p>
<pre><strong>输入：</strong>m = 7, n = 3<strong><br />输出：</strong>28</pre>
<p><strong>示例 4：</strong></p>
<pre><strong>输入：</strong>m = 3, n = 3<strong><br />输出：</strong>6</pre>
<p> </p>
<p><strong>提示：</strong></p>
<ul>
    <li><code>1 <= m, n <= 100</code></li>
    <li>题目数据保证答案小于等于 <code>2 * 10<sup>9</sup></code></li>
</ul>
<p>以下<span style="color:red">错误</span>的选项是？</p>

## aop

### before

```cpp
#include <bits/stdc++.h>
using namespace std;
```

### after

```cpp
int main()
{
    Solution sol;
    int m = 3;
    int n = 7;
    int res;

    res = sol.uniquePaths(m, n);
    cout << res;
    return 0;
}
```

## 答案

```cpp

class Solution
{
public:
    int uniquePaths(int m, int n)
    {
        vector<vector<int>> path(m, vector<int>(n, 0));
        for (int i = 0; i < n; i++)
            path[0][i] = 1;
        for (int i = 0; i < m; i++)
            path[i][0] = 1;
        for (int i = 1; i < n; i++)
            for (int j = 1; j < m; j++)
                path[j][i] = path[j - 1][i + 1] + path[j + 1][i - 1];
        return path[m - 1][n - 1];
    }
};
```
## 选项


### A

```cpp
typedef vector<int> BigInt;
class Solution
{
public:
    int uniquePaths(int m, int n)
    {
        if (m == 0 || n == 0)
            return 0;
        if (m == 1 || n == 1)
            return 1;
        int m_ = m - 1 + n - 1;
        int n_ = n - 1;
        BigInt a = fac(m_);
        int result = 0;
        for (int i = n_; i >= 1; i--)
            a = div(a, i);
        for (int i = m_ - n_; i >= 1; i--)
            a = div(a, i);
        int k = a.size() - 1;
        while (a[k] == 0)
            k--;
        for (int i = k; i >= 0; i--)
            result = result * 10 + a[i];

        return result;
    }

    BigInt fac(int n)
    {
        BigInt result;
        result.push_back(1);
        for (int factor = 1; factor <= n; ++factor)
        {
            long long carry = 0;
            for (auto &item : result)
            {
                long long product = item * factor + carry;
                item = product % 10;
                carry = product / 10;
            }
            if (carry > 0)
            {
                while (carry > 0)
                {
                    result.push_back(carry % 10);
                    carry /= 10;
                }
            }
        }
        return result;
    }
    BigInt div(BigInt a, int d)
    {
        int b = 0;
        BigInt result;
        int len = a.size();
        for (int i = len - 1; i >= 0; i--)
        {
            b = b * 10 + a[i];
            result.insert(result.begin(), b / d);
            b = b % d;
        }
        return result;
    }
};
```

### B

```cpp
class Solution
{
public:
    int uniquePaths(int m, int n)
    {
        if (m <= 0 || n <= 0)
        {
            return 0;
        }

        vector<vector<int>> dp(m + 1, vector<int>(n + 1, 0));
        for (int i = 0; i < m; i++)
        {
            dp[i][0] = 1;
        }
        for (int i = 0; i < n; i++)
        {
            dp[0][i] = 1;
        }

        for (int i = 1; i < m; i++)
        {
            for (int j = 1; j < n; j++)
            {
                dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
            }
        }
        return dp[m - 1][n - 1];
    }
};
```

### C

```cpp
class Solution
{
public:
    int uniquePaths(int m, int n)
    {
        int N = m + n - 2;
        int M = m < n ? m - 1 : n - 1;

        long ans = 1;
        for (int i = 1; i <= M; i++)
            ans = ans * (N - i + 1) / i;
        return ans;
    }
};
```