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:由于只能向两个方向走,瞬间就没有了路线迂回的烦恼,题目的难度大大的降低了。先得到到达最上边和最左边的最小路径,其他的就从上面点和左边点中选一个路径短的路径加上。

当然,也可以空间上更简单一些,只缓存一行的数据,然后一行行的更新。但我懒得写了...

#include <iostream>
#include <vector>
#include <algorithm>
#include <queue>
#include <stack>
using namespace std;

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