[Leetcode 63, Medium] Unique Paths II

最新推荐文章于 2025-07-29 14:24:32 发布

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最新推荐文章于 2025-07-29 14:24:32 发布

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分类专栏： Leetcode_medium 文章标签： leetcode C++

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本文探讨了在一个包含障碍物的网格中寻找从左上角到右下角的不同路径数量的问题。文章提供了一种使用动态规划的方法来解决该问题，并给出了 C++ 代码实现。

Problem:

Follow up for "Unique Paths":

Now consider if some obstacles are added to the grids. How many unique paths would there be?

An obstacle and empty space is marked as 1 and 0 respectively in the grid.

For example,

There is one obstacle in the middle of a 3x3 grid as illustrated below.

[
  [0,0,0],
  [0,1,0],
  [0,0,0]
]

The total number of unique paths is 2.

Note: m and n will be at most 100

Analysis:

Solutions:

C++:

    int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
        int m = obstacleGrid.size();
        int n = obstacleGrid[0].size();
        if(m == 0 && n == 0)
            return 0;
            
        if(m == 0 || n == 0) {
            return 1;
        }
            
        vector<vector<int> > num_of_paths;
        for(int row = 0; row < m; ++row) {
            vector<int> new_row;
            for(int col = 0; col < n; ++col)
                new_row.push_back(1);
                
            num_of_paths.push_back(new_row);
        }
        
        for(int i = 0; i < n; ++i) {
            if(obstacleGrid[0][i] == 1) {
                for(int j = i; j < n; ++j)
                    num_of_paths[0][j] = 0;
                    
                break;
            }
        }
        
        for(int i = 0; i < m; ++i) {
            if(obstacleGrid[i][0] == 1) {
                for(int j = i; j < m; ++j)
                    num_of_paths[j][0] = 0;
                    
                break;
            }
        }
        
        for(int i = 1; i < m; ++i) {
            for(int j = 1; j < n; ++j) {
                if(obstacleGrid[i][j] == 1)
                    num_of_paths[i][j] = 0;
                else
                    num_of_paths[i][j] = num_of_paths[i - 1][j] + num_of_paths[i][j - 1];
            }
        }
        
        return num_of_paths[m - 1][n - 1];
    }

Java:

Python: