unique paths II

public class Solution {
    public static int count;
    public int uniquePathsWithObstacles(int[][] obstacleGrid) {
        // Start typing your Java solution below
        // DO NOT write main() function
        int down = obstacleGrid.length;
        int right = obstacleGrid[0].length;
        //if(down==1&&obstacleGrid[0][0]==1)return 0;
        if(obstacleGrid[0][0]==1)return 0;
        count = 0;
        dfs(0,down-1,0,right-1,obstacleGrid);
        return count;
    }
    private void dfs(int down,int down_size,int right,int right_size,int[][] obstacleGrid){
        if(down==down_size&&right==right_size){
            count++;
        }
        if(down<down_size&&obstacleGrid[down+1][right]==0)
            dfs(down+1,down_size,right,right_size,obstacleGrid);
        if(right<right_size&&obstacleGrid[down][right+1]==0)
            dfs(down,down_size,right+1,right_size,obstacleGrid);
    }
    
}

这个大数据不能通过

public class Solution {
    public int uniquePathsWithObstacles(int[][] obstacleGrid) {
        // Start typing your Java solution below
        // DO NOT write main() function
        int down = obstacleGrid.length;
        int right = obstacleGrid[0].length;
        //if(down==1&&obstacleGrid[0][0]==1)return 0;
        if(obstacleGrid[0][0]==1)return 0;
        int[][] m = new int[down][right];
        m[0][0] = 1;
        for(int j=0;j<right;j++)
            for(int i=0;i<down;i++){
                	if(i==0&&j==0)continue;
                    if(obstacleGrid[i][j]==1){
                        m[i][j]=0;
                        continue;
                    }
                    else{
                        if(i==0&&j>0)m[i][j]=m[i][j-1];
                        else if(j==0&&i>0)m[i][j]=m[i-1][j];
                        else m[i][j]=m[i-1][j]+m[i][j-1];
                    }
        }
        return m[down-1][right-1];
    }  
}