63. Unique Paths II

最新推荐文章于 2024-07-25 09:04:50 发布

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本文探讨了在网格中加入障碍物后，唯一路径数量的变化。通过递归公式和边界条件处理，解决了该问题，并提供了求解算法的实现。

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.

相比62. Unique Paths 就是增加了障碍物，递归公式为：
if(obstacleGrid[i][j]==1)
dp[i][j]=0;
else
dp[i][j]=dp[i][j-1]+dp[i-1][j];
但是要注意下边界条件的处理。

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