Minimum Path Sum

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.

解题技巧：

该题的解法类似于Unique Paths、Unique PathsII，采用动态规划的方法，不过，该题求从起点到终点的最短路径长度。

状态转换方程为：step[i][j] = min(step[i-1][j] , step[i][j-1]) + grid[i][j]

代码：

int minPathSum(vector< vector<int> >& grid){    int m = grid.size(), n = grid[0].size();    vector< vector<int> > step(m, vector<int>(n, 0));    for(int i = 0; i < m; i ++)    {        for(int j = 0 ; j < n; j ++)        {            if(i == 0 && j == 0)            {                step[i][j] = grid[i][j];            }            else if(i == 0)            {                step[i][j] = step[i][j-1] + grid[i][j];            }            else if(j == 0)            {                step[i][j] = step[i-1][j] + grid[i][j];            }            else            {                if(step[i-1][j] < step[i][j-1])                    step[i][j] = step[i-1][j] + grid[i][j];                else                    step[i][j] = step[i][j-1] + grid[i][j];            }        }    }    return step[m-1][n-1];}