leetcode [62] Unique Paths

最新推荐文章于 2025-09-12 17:31:31 发布

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最新推荐文章于 2025-09-12 17:31:31 发布

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文章标签：数据结构与算法 c/c++

原文链接：http://www.cnblogs.com/xiaobaituyun/p/10619729.html

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。通过分析问题，给出了两种解决方案：一种是采用回溯法，但这种方法容易导致超时；另一种是使用动态规划，这种方法能够有效地计算出从网格左上角到右下角的所有唯一路径数量。

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

我一开始采用回溯法做，但是超时了。

C++超时回溯法：

class Solution {
public:
    void dfs(int m,int n,int x,int y,int&count){
        if(x>=m||x<0||y<0||y>=n) return;
        if(x==m-1&&y==n-1) count++;
        dfs(m,n,x+1,y,count);
        dfs(m,n,x,y+1,count);
    }
    int uniquePaths(int m, int n) {
        int x=0,y=0,count=0;
        dfs(m,n,x,y,count);
        return count;
    }
};

后来一想这不就是特别简单的动态规划嘛，dp[i][j]=dp[i-1][j]+dp[i][j-1],(i,j)位置上的路线总数取决于(i,j-1)和(i-1,j)位置上的路线总和。

class Solution {
private:
    int dp[105][105];
public:
    int uniquePaths(int m, int n) {
        for(int i=0;i<m;i++) dp[i][0]=1;
        for(int j=0;j<n;j++) dp[0][j]=1;
        for(int i=1;i<m;i++){
            for(int j=1;j<n;j++){
                dp[i][j]=dp[i-1][j]+dp[i][j-1];
            }
        }
        return dp[m-1][n-1];
    }
};

转载于:https://www.cnblogs.com/xiaobaituyun/p/10619729.html