62. Unique Paths [LeetCode]

最新推荐文章于 2024-11-08 13:05:39 发布

原创最新推荐文章于 2024-11-08 13:05:39 发布 · 317 阅读

0 ·

CC 4.0 BY-SA版权

LeetCode解题笔记专栏收录该内容

57 篇文章

订阅专栏

本文介绍了一种计算机器人从网格左上角到右下角唯一路径数量的算法。通过动态规划的方法，逐步计算每个格子可达路径的数量，最终得出总的路径数。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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


int uniquePaths(int m, int n){
    int *ret = (int *) malloc(n * sizeof(int));
    for (int i = 0; i < n; i++)
        ret[i] = 1;
    
    for (int i = 1; i < m; i++) {
        for (int j = 1; j < n; j++) {
            ret[j] = ret[j] + ret[j-1];
        }
    }
    int res = ret[n-1];
    free(ret);
    return res;
}