Leetcode--Minimum Path Sum

Given a m x n grid filled with non-negative numbers, find a path from top left to bottom right which minimizes the sum of all numbers along its path.

Note: You can only move either down or right at any point in time.

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 Array Dynamic Programming

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解法一：記憶深搜

class Solution {
public:
    int res=0x3f3f3f3f;
    void function(vector<vector<int> > grid,int x,int y,int sum)
    {
        if(x==grid.size()&&y==grid[0].size())
        {
            if(sum<res)
                res=sum;
            return ;
        }
        else if(x==grid.size()&&y!=grid[0].size()){
            function(grid,x,y+1,sum+grid[x-1][y]);
        }
        else if(x!=grid.size()&&y==grid[0].size())
            function(grid,x+1,y,sum+grid[x][y-1]);
        else
        {
            function(grid,x+1,y,sum+grid[x][y]);
            function(grid,x,y+1,sum+grid[x][y]);
        }
    }
    int minPathSum(vector<vector<int> > &grid) {
        int row=grid.size();
        int col=grid[0].size();
        
        if(row<1||col<1)
            return 0;
            
        function(grid,0,0,0);    
        return res;
    }
};

Status: 

Time Limit Exceeded

解法2：動態規劃

best[i][j]=min{best[i-1][j],best[i][j-1]}+grid[i][j]

class Solution {
public:
    #define MIN(a,b)  (((a)<(b))?(a):(b))
    int minPathSum(vector<vector<int> > &grid) {
        int row=grid.size();
        int col=grid[0].size();
        int res=0;
        if(row<1||col<1)
            return 0;
            
        //function(grid,0,0,0);    
        if(row==1)
        {
            for(int i=0;i<col;i++)
                res+=grid[0][i];
            return res;
        }
        
        if(col==1)
        {
            for(int i=0;i<row;i++)
                res+=grid[i][0];
            return res;
        }
        int **best=new int*[row];
        for(int i=0;i<row;i++)
        {
            best[i]=new int[col];
            memset(best[i],0,sizeof(int)*col);
        }
       // best[0][0]=grid[0][0];

        for(int i=0;i<row;i++)
            for(int j=0;j<col;j++)
            {
                if(i==0&&j==0)
                    best[0][0]=grid[0][0];
                else if(i==0&&j!=0)
                    best[i][j]=best[i][j-1]+grid[i][j];
                else if(i!=0&&j==0)
                    best[i][j]=best[i-1][j]+grid[i][j];
                else
                    best[i][j]=MIN(best[i-1][j],best[i][j-1])+grid[i][j];
            }
        return best[row-1][col-1];
    }
};