问题

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？

例如，上图是一个7 x 3 的网格。有多少可能的路径？

说明：m 和 n 的值均不超过 100。

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。

向右 -> 向右 -> 向下
向右 -> 向下 -> 向右
向下 -> 向右 -> 向右

输入: m = 7, n = 3

输出: 28

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.

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:

Right -> Right -> Down
Right -> Down -> Right
Down -> Right -> Right

Input: m = 7, n = 3

Output: 28

示例

public class Program {

    public static void Main(string[] args) {
        var m = 7;
        var n = 3;

        var res = UniquePaths(m, n);
        Console.WriteLine(res);

        Console.ReadKey();
    }

    private static int UniquePaths(int m, int n) {
        if(m == 0 || n == 0) return 0;
        var dp = new int[m, n];
        for(var i = 0; i < m; i++) {
            for(var j = 0; j < n; j++) {
                dp[i, j] = 1;
            }
        }
        for(var i = 1; i < m; i++) {
            for(var j = 1; j < n; j++) {
                dp[i, j] = dp[i, j - 1] + dp[i - 1, j];
            }
        }
        return dp[m - 1, n - 1];
    }

}

以上给出1种算法实现，以下是这个案例的输出结果：