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.

思路：簡單動態規劃，考慮用 vector 代替陣列，來節省空間。

//陣列版，13ms
class Solution {
public:
    int minPathSum(vector
 
<vector<int>>& grid) {
        int m = grid.size(), n = grid[0].size();
        long long dp[300][300];
        dp[0][0] = grid[0][0];
        //初始化邊界
        for(int i=1;i<m;i++)
            dp[i][0] = dp[i-1][0]+grid[i][0];
        for(int j=1;j<n;j++)
            dp[0][j] = dp[0][j-1]+grid[0
 
][j];
        //處理中間部分
        for(int i=1;i<m;i++)
        {
            for(int j=1;j<n;j++)
            {
                dp[i][j] = min(dp[i-1][j],dp[i][j-1]) + grid[i][j];
            }
        }

        return dp[m-1][n-1];
    }
};

//vector版，9ms
class Solution {
public:
    int minPathSum(vector
 
<vector<int>>& grid) {
        int m = grid.size(), n = grid[0].size();
        vector<vector<int> > dp;
        vector<int> line0;
        line0.push_back(grid[0][0]);
        dp.push_back(line0);
        //初始化邊界
        for(int i=1;i<m;i++)
        {
            vector<int> tmp;
            tmp.push_back(dp[i-1][0]+grid[i][0]);
            dp.push_back(tmp);
        }

        for(int j=1;j<n;j++)
            dp[0].push_back(dp[0][j-1]+grid[0][j]);
        //處理中間部分
        for(int i=1;i<m;i++)
        {
            for(int j=1;j<n;j++)
            {
                dp[i].push_back(min(dp[i-1][j],dp[i][j-1]) + grid[i][j]);
            }
        }

        return dp[m-1][n-1];
    }
};