[leetcode] 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.

解法一：

recursion的思路，不过通过不了OJ的runtime test。因为很多grid的node走了太多次了。。。

class Solution {public:    int minPathSum(vector<vector<int>>& grid) {        int res = 0;        if(grid.empty() || grid[0].empty()) return res;        res += grid[0][0];                if(grid.size()==1){            for(int i=1; i<grid[0].size(); i++)                res += grid[0][i];            return res;        }                if(grid[0].size()==1){            for(int i=1; i<grid.size(); i++)                res += grid[i][0];            return res;        }                vector<vector<int>> down_grid, right_grid;        for(int i=1; i<grid.size(); i++){            down_grid.push_back(grid[i]);        }        for(int i=0; i<grid.size(); i++){            vector<int> tmp = grid[i];            tmp.erase(tmp.begin());            right_grid.push_back(tmp);        }                int sum_down = minPathSum(down_grid);        int sum_right = minPathSum(right_grid);        res += min(sum_down, sum_right);        return res;            }};

解法二：

还可以用dp的方法去做，复杂度显然就是O(m*n)了。用一个二维数组dp，dp[i][j]表示走到i,j时最小的sum，很容易得出递推公式是dp[i][j] = grid[i][j] + min(dp[i-1][j], dp[i][j-1])。

class Solution {public:    int minPathSum(vector<vector<int>>& grid) {        if(grid.empty() || grid[0].empty()) return 0;                int m = grid.size(), n= grid[0].size();        int dp[m][n];        dp[0][0] = grid[0][0];        for(int i=1;i<m; i++) dp[i][0]=grid[i][0]+dp[i-1][0];        for(int j=1;j<n; j++) dp[0][j]=grid[0][j]+dp[0][j-1];                for(int i=1; i<m; i++)            for(int j=1; j<n; j++){                dp[i][j] = grid[i][j] + min(dp[i-1][j], dp[i][j-1]);            }        return dp[m-1][n-1];                    }};