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];}