Unique Paths II-优快云博客

本文链接：https://blog.youkuaiyun.com/caijhBlog/article/details/79117810

本文详细解析了UniquePathsII算法，该算法为UniquePaths的拓展，考虑了网格中存在障碍物的情况，通过动态规划求解从起点到终点的不同路径数量。文章提供了C++代码实现，并解释了关键步骤。

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Unique Paths II

A.题意

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,
[
[0,0,0],
[0,1,0],
[0,0,0]
]
The total number of unique paths is 2.
There is one obstacle in the middle of a 3x3 grid as illustrated below.
Note: m and n will be at most 100.

这道题是前面提到过的Unique Path的拓展，这里遇到1就不能走过去了，还是和之前一样让你求总的方法数

B.思路

这道题目其实就遇到1时那一个格子方法数置0，递推公式一样，然后初始化的时候注意前面是不是有1，有的话后面全部置0.

C.代码实现

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