63. Unique Paths II

最新推荐文章于 2024-07-25 09:04:50 发布

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本文探讨了在网格中加入障碍物后，唯一路径的数量如何变化。通过实例展示了如何计算在存在障碍物的3x3网格中，唯一路径总数为2的方法。

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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.

int uniquePathsWithObstacles(vector<vector<int>>& o) {
    int m = o.size();
    int n = o[0].size();
    vector<vector<int>> res(m, vector<int>(n, 1));
    for (int i = 0; i < m; i++) {
        for (int j = 0; j < n; j++) {
            if (o[i][j])
                res[i][j] = 0;
            else if (i == 0 && j == 0)
                res[i][j] = 1;
            else {
                int up = i == 0 ? 0 : res[i - 1][j];
                int left = j == 0 ? 0 : res[i][j - 1];
                res[i][j] = up + left;
            }      
        }
    }
    return res[m - 1][n - 1];
}