[LeetCode]Unique Paths II

最新推荐文章于 2025-01-13 17:23:47 发布

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最新推荐文章于 2025-01-13 17:23:47 发布

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分类专栏： LeetCode

本文链接：https://blog.youkuaiyun.com/sunbaigui/article/details/8981353

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本文介绍了一种使用动态规划解决带有障碍物的网格中从起点到终点的不同路径数量的方法。通过两个不同的C++实现展示了如何根据网格中是否存在障碍物来更新路径计数。

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class Solution {
//DP
//if a[i][j] is a obstacle, then f[i][j]=0
public:
	 int uniquePathsWithObstacles(vector<vector<int> > &obstacleGrid) {
		// Start typing your C/C++ solution below
		// DO NOT write int main() function
		 int m = obstacleGrid.size();
		 if(0 == m) return 0;
		 int n = obstacleGrid[0].size();
		vector<vector<int>> f(m+1, vector<int>(n+1, 0));
		if(obstacleGrid[0][0] != 1)
			f[1][1] = 1;
		for (int i = 1; i <= m; ++i)
		{
			for (int j = 1; j <= n; ++j)
			{
				if(i == 1 && j == 1) continue;
				if(0 == obstacleGrid[i-1][j-1])//empty 
					f[i][j] = f[i-1][j]+f[i][j-1];
			}
		}
		return f[m][n];
	}
};

second time

class Solution {
public:
    int uniquePathsWithObstacles(vector<vector<int> > &obstacleGrid) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        int m = obstacleGrid.size();
        if(m == 0) return 0;
        int n = obstacleGrid[0].size();
        
        vector<vector<int> > f(m+1, vector<int>(n+1, 0));
        for(int i = 1; i <= m; ++i)
        {
            for(int j = 1; j <= n; ++j)
            {
                if(obstacleGrid[i-1][j-1] == 1) continue;
                
                if(i == 1 && j == 1) f[i][j] = 1;
                else f[i][j] = f[i-1][j]+f[i][j-1];
            }
        }
        return f[m][n];
    }
};