119-Minimum Path Sum

-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.

思路：动态规划
d[i][j]=grid[i][j]+min(d[i−1][j],d[i][j−1])
(处理边界，第一行和第一列为累积和)

class Solution {public:    int minPathSum(vector<vector<int>>& grid) {        int row = grid.size();        if(row==0)return 0;        int col = grid[0].size();        vector<int> v(col,0);        vector<vector<int>> d(row,v);        int row_sum =0;        for(int i=0;i<col;++i){            row_sum += grid[0][i];            d[0][i] = row_sum;        }        row_sum =0;        for(int i=0;i<row;++i){            row_sum += grid[i][0];            d[i][0] = row_sum;        }        //动态规划解决，滚动数组        for(int i=1;i<row;++i){            for(int j=1;j<col;++j){                d[i][j] = grid[i][j]+min(d[i-1][j],d[i][j-1]);            }        }        return d[row-1][col-1];    }};

递归版，超时

class Solution {public:    int minPathSum(vector<vector<int>>& grid) {        int row = grid.size();        if(row==0)return 0;        int col = grid[0].size();        vector<vector<int>> d= grid;        int row_sum =0;        for(int i=0;i<col;++i){            row_sum += grid[0][i];            d[0][i] = row_sum;        }        row_sum =0;        for(int i=0;i<row;++i){            row_sum += grid[i][0];            d[i][0] = row_sum;        }        return minPathSum_Recursive(grid,row-1,col-1);    }    int minPathSum_Recursive(vector<vector<int>> &d,int i,int j){        if(i==0||j==0){            return d[i][j];        }        else{            int res;            res = d[i][j]+min(minPathSum_Recursive(d,i-1,j),minPathSum_Recursive(d,i,j-1));            return res;        }    }   };