Unique Paths II

最新推荐文章于 2024-07-25 09:04:50 发布

原创最新推荐文章于 2024-07-25 09:04:50 发布 · 401 阅读

0 ·

CC 4.0 BY-SA版权

算法专栏收录该内容

636 篇文章

订阅专栏

本文探讨了在网格中加入障碍物后，独特的路径数量如何计算。通过实例展示了一个包含障碍物的3x3网格，并说明了在该网格中唯一路径的数量为2。详细介绍了算法实现过程，包括初始化缓冲矩阵、标记起点和终点、遍历网格并更新路径计数。

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.

class Solution {
public:
    int uniquePathsWithObstacles(vector<vector<int> > &obstacleGrid) {
        int m = obstacleGrid.size();
        if (m < 1)
        {
            return 0;
        }
        int n = obstacleGrid[0].size();
        if (n < 1)
        {
            return 0;
        }

        int buf[m][n];
        bool obstacleFound = false;
        for (int i = 0; i < m; i++)
        {
            if (!obstacleFound && obstacleGrid[i][0] == 0)
            {
                buf[i][0] = 1;
            }
            else
            {
                buf[i][0] = 0;
                obstacleFound = true;
            }
        }

        obstacleFound = false;
        for (int i = 0; i < n; i++)
        {
            if (!obstacleFound && obstacleGrid[0][i] == 0)
            {
                buf[0][i] = 1;
            }
            else
            {
                buf[0][i] = 0;
                obstacleFound = true;
            }
        }

        for (int i = 1; i < m; i++)
        {
            for (int j = 1; j < n; j++)
            {
                if (obstacleGrid[i][j] == 1)
                {
                    buf[i][j] = 0;
                }
                else
                {
                    buf[i][j] = buf[i-1][j] + buf[i][j-1];
                }
            }
        }

        return buf[m-1][n-1];
    }
};