64. 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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思路：经典DP动态规划入门问题，

定义状态转移方程f[i][j] = min(f[i-1][j],f[i][j-1])+grid[i][j]

初始状态函数，二维数组f[][]的第一行和第一列

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code 如下：

class Solution{
public:
    int minPathSum(vector<vector<int> > &grid){
        if(grid.size()==) return ;
        const int m = grid.size();
        const int n = grid[].size();

        int f[m][n];
        f[][] = grid[][];
        for(int i = ;i<m;i++){
            //初始化状态方程第一lie
            f[i][] = f[i-][]+grid[i][];
        }
        for(int i = ;i<n;i++){
            //初始化状态方程第一hang
            f[][i] = f[][i-]+grid[][i];
        }

        //运用状态方程求解
        for(int i = ;i<m;i++){
            for(int j = ;j<n;j++){
                if(f[i-][j]<f[i][j-]){
                    cout<<i-<<"-"<<j<<endl;
                }else{
                    cout<<i<<"-"<<j-<<endl;
                }
                f[i][j] = min(f[i-][j],f[i][j-])+grid[i][j];
            }
        }

        return f[m-][n-];
    }
};