LeetCode OJ - 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.

分析：动态规划，递归和迭代都行

class Solution {    int dp[1000][1000];public:    int DFS(vector<vector<int> > &grid, int i, int j) {        if(i < 0 || j < 0) return INT_MAX;                if(dp[i][j] != -1) {            return dp[i][j];        }            dp[i][j] =  min( DFS(grid, i - 1, j) ,  DFS(grid, i, j - 1) ) + grid[i][j];        return dp[i][j];    }        int minPathSum(vector<vector<int> > &grid) {        int m = grid.size();        int n = grid[0].size();        for(int i = 0; i < m; i++) {            for(int j = 0; j < n; j++) {                    dp[i][j] = -1;            }        }                dp[0][0] = grid[0][0];        return DFS(grid, m - 1, n - 1);    }};

class Solution {    int dp[1000][1000];public:    int minPathSum(vector<vector<int> > &grid) {        int m = grid.size();        int n = grid[0].size();                int i, j;        int path = 0;        for(i = 0; i < n; i++) {            dp[0][i] = path + grid[0][i];            path = dp[0][i];        }                path = 0;        for(j = 0; j < m; j++) {            dp[j][0] = path + grid[j][0];            path = dp[j][0];        }                int ret = INT_MAX;        for(i = 1; i < m; i++) {            for(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];    }};