[leetcode] Unique Paths II

最新推荐文章于 2019-05-17 12:08:12 发布

原创最新推荐文章于 2019-05-17 12:08:12 发布 · 554 阅读

0 ·

CC 4.0 BY-SA版权

算法题专栏收录该内容

107 篇文章

订阅专栏

本文探讨了如何使用动态规划方法解决在网格中包含障碍物时的唯一路径问题。通过实例分析，详细解释了算法实现过程，并提供了关键代码片段。适合对动态规划和网格问题解决感兴趣的读者。

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.

同样的动态规划走起：

class Solution {
public:
    int uniquePathsWithObstacles(vector<vector<int> > &obstacleGrid) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        int m=obstacleGrid.size();
        int n=0;
        if(m)
            n=obstacleGrid[0].size();
        if(m==0 || n==0)
        	return 0;
        int **p=new int*[m];
        for(int i=0 ; i<m ; i++)
        	p[i]=new int[n];
        p[0][0] = obstacleGrid[0][0] == 1 ? 0 : 1;
        for(int i = 1; i < m; i++)
            p[i][0] = obstacleGrid[i][0] == 1 ? 0 : p[i-1][0];
            
        for(int i = 1; i < n ; i++)
            p[0][i] = obstacleGrid[0][i] == 1 ? 0 : p[0][i-1];
            
        for(int i = 1; i < m; i++)
            for(int j = 1; j < n; j++)
                p[i][j] = obstacleGrid[i][j] == 1 ? 0 : p[i-1][j] + p[i][j-1];
                
        return p[m-1][n-1];
    }
};