LeetCode 题解（50）: Unique Paths

最新推荐文章于 2025-09-10 22:03:14 发布

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最新推荐文章于 2025-09-10 22:03:14 发布

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分类专栏：算法文章标签： algorithm leetcode

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本文介绍了一个经典的动态规划问题——计算机器人从网格左上角到右下角的不同路径数量。提供了C++, Java及Python三种语言的实现代码。

题目：

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.

题解：动态规划。

C++版：

class Solution {
public:
    int uniquePaths(int m, int n) {
        if(!m || !n)
            return 0;
        if(m == 1 || n == 1)
            return 1;
        
        vector<vector<int>> ways;
        for(int i = 0; i < m; i++) {
            vector<int> way(n);
            ways.push_back(way);
        }
        
        for(int j = 0; j < n; j++)
            ways[0][j] = 1;
        for(int k = 0; k < m; k++)
            ways[k][0] = 1;
            
        for(int x = 1; x < m; x++) 
            for(int y = 1; y < n; y++) 
                ways[x][y] = ways[x][y-1] + ways[x-1][y];
        
        return ways[m-1][n-1];
    }
};

Java版：

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

Python版：

class Solution:
    # @return an integer
    def uniquePaths(self, m, n):
        if m == 0 or n == 0:
            return 0
        if m == 1 or n == 1:
            return 1
        paths = []
        paths.append([1] * n)
        for i in range(1, m):
            temp = [0] * (n-1)
            temp.insert(0, 1)
            paths.append(temp)
            
        for j in range(1, m):
            for k in range(1, n):
                paths[j][k] = paths[j-1][k] + paths[j][k-1]
                
        return paths[m-1][n-1]