leetcode unique path I&&II

I

动态规划，一次性构造，查询。采用回溯法会超时

class Solution {
public:
int grid[100][100];
    int uniquePaths(int m, int n) 
    {
       if(grid[m-1][n-1]==0) construct();
       return grid[m-1][n-1];
    }
    void construct()
    {
        grid[0][0]=1;
        for(int i=1;i<100;i++)
        {
           grid[i][0]=1; 
        }
        for(int i=1;i<100;i++)
        {
           grid[0][i]=1; 
        }
        for(int i=1;i<100;i++)
        {
            for(int j=1;j<=i;j++)
            {
                grid[i][j]=grid[i][j-1]+grid[i-1][j];
                grid[j][i]=grid[i][j];
            }
        }
    }       
};

 II

有障碍物的情况

class Solution {
public:
int grid[100][100];
    int uniquePathsWithObstacles(vector<vector<int> > &obstacleGrid) 
    {
        int m=obstacleGrid.size();
        if(m==0)return 0;
        int n=obstacleGrid[0].size();
        for(int i=0;i<m;i++)
        {
            for(int j=0;j<n;j++)
            grid[i][j]=0;
        }
        if(!obstacleGrid[0][0])
        grid[0][0]=1;
        for(int i=1;i<m;i++)
        {
           if(!obstacleGrid[i][0])
           grid[i][0]=grid[i-1][0];
        }
        for(int i=1;i<n;i++)
        {
           if(!obstacleGrid[0][i])
           grid[0][i]=grid[0][i-1]; 
        }
        for(int i=1;i<m;i++)
        {
            for(int j=1;j<n;j++)
            {
                if(!obstacleGrid[i][j])
                {
                    grid[i][j]+=grid[i-1][j]+grid[i][j-1];  
                }                                           
            }
        }
        return grid[m-1][n-1];        
    }
};