LeetCode实战：不同路径

题目英文

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

题目中文

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。

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

示例 3:

输入: m = 23, n = 12
输出: 193536720

算法实现

方式一：利用递归

public class Solution
{private int _m;private int _n;public int UniquePaths(int m, int n){_m = m;_n = n;int count = 0;RecordPaths(0, 0, ref count);return count;}private void RecordPaths(int i, int j, ref int count){if (i == _m - 1 && j == _n - 1){count++;return;}if (i < _m){RecordPaths(i + 1, j, ref count);}if (j < _n){RecordPaths(i, j + 1, ref count);}}
}

使用递归的方式，容易理解但会耗费大量的时间，所以在运行示例3 的时候，超时了。

方式二：利用动态规划

动态规划表格01：

表01

动态规划表格02：

表02

动态规划的最优子结构为：d[i,j] = d[i-1,j] + d[i,j-1]

public class Solution
{public int UniquePaths(int m, int n){int[,] memo = new int[m, n];for (int i = 0; i < m; i++){for (int j = 0; j < n; j++){if (i == 0){memo[i, j] = 1;}else if (j == 0){memo[i, j] = 1;}else{memo[i, j] = memo[i - 1, j] + memo[i, j - 1];}}}return memo[m - 1, n - 1];}
}