1、题目

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?

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

2、分析

及其经典的动态规划题。这类题目做的太多了，思路十分明确。机器人只能往右边和下边走。那么机器人到达一个位置，只能是从该位置的上面或者左边过来。假如到达该位置的上面有a种方法，到达该位置的左面有b种方法，那么到达该位置就有a+b种方法。递推就出来了。用dp数组来表示。dp[i][j]为到达第i行第j列的位置的方法数。一开始的时候第一行和第一列要初始化为1.然后递推就是dp[i][j] = dp[i-1][j]+dp[i][j-1]。注意只有一个格子的特殊情况即可。

3、代码

class Solution {
public:
    int uniquePaths(int m, int n) {
        int dp[m][n] = {0};
        dp[0][0]=1;
        for(int i=1;i<m;i++) dp[i][0] = 1;
        for(int i=1;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];
        
    }
};