leetCode+Unique Paths II

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,
ere is one obstacle in the middle of a 3 × 3 grid as illustrated below.
[
[0,0,0],
[0,1,0],
[0,0,0]
]
e total number of unique paths is 2.
Note: m and n will be at most 100.

方法一：DFS

class Solution {
public:
    int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
        if(obstacleGrid.empty()) return 0;

        int m = obstacleGrid.size();
        int n = obstacleGrid[0].size();
        if(obstacleGrid[0][0] || obstacleGrid[m-1][n-1]) return 0;

        int **buffer = new int* [m+1];
        for(int i=0;i<=m;i++){
            buffer[i] = new int [n+1];
            fill(buffer[i],buffer[i]+n+1,0);
        }
        int temp = dfs(obstacleGrid,m,n,buffer);
        for(int i=0;i<=m;i++)
            delete [] buffer[i];
        delete [] buffer;
        return temp;
    }

    int dfs(vector<vector<int>> &obstacleGrid,int x,int y,int** buf){
        if(x < 1 || y < 1) return 0;
        if(obstacleGrid[x-1][y-1]) return 0;
        if(buf[x][y]>0) return buf[x][y];
        if(x == 1 && y==1 ) return 1;

        buf[x-1][y] = dfs(obstacleGrid, x-1, y, buf); 
        buf[x][y-1] = dfs(obstacleGrid, x, y-1, buf);
        return buf[x][y] = buf[x-1][y]+buf[x][y-1];
    }
};

方法二：DP

class Solution {
public:
    int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
        if(obstacleGrid.empty()) return 0;

        int m = obstacleGrid.size();
        int n = obstacleGrid[0].size();
        if(obstacleGrid[0][0] || obstacleGrid[m-1][n-1]) return 0;
        int *buf = new int [n];
        fill(buf,buf+n,0);

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