leetcode Minimum Path Sum

题目：还是类似于之前两题，这里给定的m*n矩阵是每个数字都有值的。求的从头到尾的最小sum是多少。还是用动态规划求。初始化的时候是先计算第一行和第一列。假设给定的数据如下：

那么初始化后为：

那么再定位要处理的数字到i=1和j=1处，如表中的数字2处。这个时候更新是更加它加上它的左边或者上边的较小的值。更新一次后为：

一次类推可以将所有的更新完，最后输出grid[m-1][n-1]就可以了。

class Solution {
public:
    int minPathSum(vector<vector<int> > &grid) 
    {
        //if (grid.size() == 0) return 0;
        int m = grid.size(), n = grid[0].size();
        int i =1, j = 1; 
        while(j<n) { grid[0][j]+=grid[0][j-1]; j++;}
        while(i<m) { grid[i][0]+=grid[i-1][0]; i++;}
        i = 1; j = 1;
        for (; i < m; ++i)
            for (j =1 ; j < n; ++j) // 记得j重新赋值1
            {
                grid[i][j] += min(grid[i-1][j], grid[i][j-1]);
            }
        return grid[m-1][n-1];
    }
};

 2015/03/25:

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

python:

class Solution:
    # @param grid, a list of lists of integers
    # @return an integer
    def minPathSum(self, grid):
        for i in range(1, len(grid)):
            grid[i][0] += grid[i-1][0]
        for i in range(1, len(grid[0])):
            grid[0][i] += grid[0][i-1]
        for i in range(1, len(grid)):
            for j in range(1, len(grid[0])):
                grid[i][j] = min(grid[i-1][j], grid[i][j-1]) + grid[i][j]
        return grid[len(grid)-1][len(grid[0])-1]