62. Unique Paths

最新推荐文章于 2025-04-02 22:31:45 发布

转载最新推荐文章于 2025-04-02 22:31:45 发布 · 77 阅读

本文介绍了一个经典的算法问题：计算从网格左上角到右下角的不同路径数量，机器人只能向右或向下移动。通过动态规划的方法，展示了如何高效地解决这个问题。

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

class Solution {
public:
    int uniquePaths(int m, int n) {
        vector<vector<int> > path(m, vector<int> (n, 1));
        for (int i = 1; i < m; i++)
            for (int j = 1; j < n; j++)
                path[i][j] = path[i - 1][j] + path[i][j - 1];
        return path[m - 1][n - 1];
    }
};