62. Unique Paths

最新推荐文章于 2025-04-02 22:31:45 发布

_风铃

最新推荐文章于 2025-04-02 22:31:45 发布

阅读量144

点赞数

分类专栏：算法分析与设计

本文链接：https://blog.youkuaiyun.com/wuchg6/article/details/78489479

版权

算法分析与设计专栏收录该内容

21 篇文章

订阅专栏

本文介绍了一种使用动态规划解决机器人网格路径计数问题的方法。通过构建有向图并定义递推公式 C(i,j) = C(i-1,j) + C(i,j-1)，解决了从网格左上角到右下角的所有唯一路径数量的问题。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Description:

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 3 x 7 grid. How many possible unique paths are there?

Note: m and n will be at most 100.

简要题解：

采用动态规划。将grid抽象为有向图，每一个格子为图中的一个结点，图中包含所有向下和向右的边。令C(i, j)为以grid[0][0]为起点，grid[i][j]为终点的路径个数（i或j小于0时，C(i, j) = 0）。则C(0, 0) = 1，并且，一般地，C(i, j) = C(i - 1, j) + C(i, j - 1)，便可求出从起点到终点的路径个数。

代码：

class Solution {
public:
    int uniquePaths(int m, int n) {
        vector<vector<int> > grid;

        grid.resize(m);
        for (int i = 0; i < m; i++)
            grid[i].resize(n);

        for (int i = 0; i < m; i++)
            for (int j = 0; j < n; j++)
                grid[i][j] = 0;

        grid[0][0] = 1;

        for (int i = 0; i < m; i++)
            for (int j = 0; j < n; j++) {
                if (i > 0)
                    grid[i][j] += grid[i-1][j];
                if (j > 0)
                    grid[i][j] += grid[i][j-1];
            }
        
        return grid[m-1][n-1];
    }
};