[LeetCode] Unique Paths II

hale1007

于 2014-10-23 13:39:16 发布

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分类专栏： Algorithm LeetCode

本文链接：https://blog.youkuaiyun.com/hale1007/article/details/40395681

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探讨了在网格中加入障碍物后，不同路径数量的变化。通过实例展示了一个3x3网格中，存在一个障碍物时，唯一路径的数量为2。代码实现了一个解决方案类，用于计算带有障碍物的网格中的独特路径数。

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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.

class Solution {
public:
    int uniquePathsWithObstacles(vector<vector<int> > &obstacleGrid) {
        int m = obstacleGrid.size(), n = obstacleGrid[0].size();
        vector<vector<int>> f(m, vector<int>(n, 0));
        for(int i = 0; i < m; i++)
        {
            if(obstacleGrid[i][0] == 0)
                f[i][0] = 1;
            else
                break;
        }
        for(int i = 0; i < n; i++)
        {
            if(obstacleGrid[0][i] == 0)
                f[0][i] = 1;
            else
                break;
        }
        for(int i = 1; i < m; i++)
        {
            for(int j = 1; j < n; j++)
            {
                if(obstacleGrid[i][j] == 0)
                    f[i][j] = f[i-1][j] + f[i][j-1];
                else
                    f[i][j] = 0;
            }
        }
        return f[m-1][n-1];
    }
};