leetcode---Unique Paths II---动规

最新推荐文章于 2021-11-14 23:25:40 发布

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最新推荐文章于 2021-11-14 23:25:40 发布

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本文介绍了一个基于动态规划算法的问题——在包含障碍物的网格中计算从起点到终点的不同路径数量。文章提供了一种有效的实现方法，并通过代码示例详细解释了如何避免障碍并计算可行路径。

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) 
    {
        int row = obstacleGrid.size();
        int col = obstacleGrid[0].size();
        int dp[101][101];
        memset(dp, 0, sizeof(dp));
        dp[0][0] = obstacleGrid[0][0] == 0? 1 : 0;
        for(int i=1; i<row; i++)
            if(obstacleGrid[i][0] == 0)
                dp[i][0] = dp[i-1][0];
        for(int i=1; i<col; i++)
            if(obstacleGrid[0][i] == 0)
                dp[0][i] = dp[0][i-1];
        for(int i=1; i<row; i++)
            for(int j=1; j<col; j++)
                dp[i][j] = obstacleGrid[i][j]==0? dp[i-1][j] + dp[i][j-1] : 0;
        return dp[row-1][col-1];
    }
};