【Leetcode】Unique Paths

最新推荐文章于 2025-09-11 23:50:21 发布

转载最新推荐文章于 2025-09-11 23:50:21 发布 · 121 阅读

文章标签：

#java #leetcode #array

本文探讨了机器人从网格地图的左上角到达右下角的路径问题，通过动态规划算法计算可能的独特路径数量。重点介绍了两种动态规划方法：一种使用二维数组，另一种采用一维数组以节省内存。实例分析了3x7网格的情况。

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.

Java:

1. 动态规划： http://blog.youkuaiyun.com/lilong_dream/article/details/19771225 时不时会超时

public class Solution {  
    public int uniquePaths(int m, int n) {  
        int[][] A = new int[m][n];  
  
        for (int i = 0; i < m; ++i) {  
            A[i][0] = 1;  
        }  
  
        for (int i = 1; i < n; ++i) {  
            A[0][i] = 1;  
        }  
  
        for (int i = 1; i < m; ++i)  
            for (int j = 1; j < n; ++j) {  
                A[i][j] = A[i][j - 1] + A[i - 1][j];  
            }  
  
        return A[m - 1][n - 1];  
    }

2. 省内存的动态规划：http://blog.youkuaiyun.com/linhuanmars/article/details/22126357

public int uniquePaths(int m, int n) {  
    if(m<=0 || n<=0)  
        return 0;  
    int[] res = new int[n];  
    res[0] = 1;  
    for(int i=0;i<m;i++)  
    {  
        for(int j=1;j<n;j++)  
        {  
           res[j] += res[j-1];  
        }  
    }  
    return res[n-1];  
}