标签：style blog io color os sp for on 数据
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(); } };
【leetcode】Minimum Path Sum(easy)
标签：style blog io color os sp for on 数据
原文地址：http://www.cnblogs.com/dplearning/p/4153597.html
