62. Unique Paths

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本文探讨了一个位于m x n网格上的机器人如何从左上角移动到右下角的问题，机器人只能向下或向右移动。通过动态规划算法，我们计算了到达目标的所有唯一路径数量。文章详细解释了算法的实现过程，并提供了代码示例。

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62. Unique Paths

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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

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class Solution {
public:
    int uniquePaths(int m, int n) {
        int dp[n][m]={};//n is row and m is col
        for(int i=0;i<n;i++) dp[i][0]=1;
        for(int i=0;i<m;i++) dp[0][i]=1;
        for(int i=1;i<n;i++){
        	for(int j=1;j<m;j++){
        		dp[i][j]=dp[i-1][j]+dp[i][j-1];
        	}
        }
        return dp[n-1][m-1];
    }

};