Unique Paths II

最新推荐文章于 2025-01-13 17:23:47 发布

原创最新推荐文章于 2025-01-13 17:23:47 发布 · 455 阅读

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本文探讨了在一个含有障碍物的网格中寻找从起点到终点的不同路径数量的问题。通过动态规划的方法，给出了一个高效的解决方案。

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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) {
        size_t m = obstacleGrid.size();
        if(!m) return 0;
        size_t n = obstacleGrid[0].size();
        vector<vector<int>> paths(m,vector<int>(n,0));
        for(size_t i=0;i<obstacleGrid.size();++i){
            for(size_t j=0;j<obstacleGrid[i].size();++j){
                if(i==0 && j==0) {
                    if(obstacleGrid[i][j]) break;
                    paths[i][j] = 1;
                    continue;
                }
                if(obstacleGrid[i][j]) continue;
                if(i>0 && j>0) paths[i][j] = paths[i-1][j]+paths[i][j-1];
                else if(i>0) paths[i][j] = paths[i-1][j];
                else paths[i][j] = paths[i][j-1];
            }
        }
        return paths[m-1][n-1];
    }
};