[LeetCode] 62. Unique Paths

Description

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

Analyse

有一个(m imes n)的网格，机器人处于左上方(0, 0)位置处，机器人只能向右或者向下走，计算一个机器人从(0, 0)走到(m, n)的总路线数

动态规划

dp[i][j] 代表从(0, 0)走到(i, j)的总路线数

机器人只能向右或者向下走，所以当机器人在(i, j)的时候，有两种情况
从(i-1, j)向下走了一步
从(i, j-1)向右走了一步

dp[i][j] = dp[i, j-1] + dp[i-1, j]

int uniquePaths(int m, int n) {
    if (m < 0 || n < 0) return 0;
    int dp[100][100];

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

    for (int j = 0; j < n; j++) {
        dp[0][j] = 1;
    }

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

    return dp[m-1][n-1];
}

组合数学

从(0, 0)走到(m ,n)，需向右走n-1步，向下走m-1步，共 m+n-2步，路径数位(C_{m+n-2}^{m-1}) 或 (C_{m+n-2}^{n-1})

[C_{m+n-2}^{n-1} = frac{(m+n-2)!}{(m-1)!(n-1)!} ]

[C_{m+n-2}^{m-1} = frac{(m+n-2)!}{(n-1)!(m-1)!} ]

(C_m^n)的计算方法

[C_m^n = frac{m!}{n!(m-n)!} ]

[C_4^2 = frac{4 imes 3}{2 imes 1} = frac{4}{2} imes frac{3}{1} ]

用编程实现时可以用n控制循环次数，循环n次，依次乘上 (frac{m - i}{n - i})，i为循环变量

下面这段代码看似正确，实际上无法通过，原因是做除法的时候可能无法整除，导致计算错误

int combination(int m, int n) {
    unsigned long long ans = 1;

    for(int i = 0; i < n; i++) {
        ans *= (m - i);
        ans /= (n - i); // 此处有错误
    }

    return ans;
}
int uniquePaths(int m, int n) {
    int a = m + n -2;
    int b = min(n-1, m-1); // 利用组合的性质减少计算次数
    return combination(a, b);
}

通过更换除数的顺序解决无法整除的问题， 最终代码如下

 int combination(int m, int n) {
    unsigned long long ans = 1;

    for(int i = 0; i < n; i++) {
        ans *= (m - i);
        ans /= (i + 1); // 从1开始除，可以保证能够整除
    }

    return ans;
}
int uniquePaths(int m, int n) {
    int a = m + n -2;
    int b = min(n-1, m-1);
    return combination(a, b);
}