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：本题的解题思路与《Unique Paths》一样，也是典型的动态规划问题。要求从左上角grid[0][0]到右下角grid[m-1][n-1]的路径最小值，可以从grid[m-1][n-1]开始，从右向左，从下到上的依次求出grid[i][j]到grid[m-1][n-1]的路径最小值。

        设min[i][j]表示从grid[i][j]到grid[m-1][n-1]的最小值。因只能向下或者向右走，所以min[i][j] = grid[i][j]+min(min[i][j+1],min[i+1][j])，因此，只要求出了min[i][j+1]和min[i+1][j]，就可以求出min[i][j]了。代码如下：

int minPathSum(int** grid, int gridRowSize, int gridColSize) 
{
    int i, j;
    for(i = gridRowSize-1; i >= 0; i--)
    {
        for(j = gridColSize-1; j >= 0; j--)
        {
            if(i == gridRowSize-1 && j == gridColSize-1)    continue;
        
            if(i == gridRowSize-1)
            {
                grid[i][j] += grid[i][j+1];
                continue;
            }

            if(j == gridColSize-1)
            {
                grid[i][j] += grid[i+1][j];
                continue;
            }

            grid[i][j] += min(grid[i+1][j], grid[i][j+1]);
        }
    }

    return grid[0][0];
}