此题不难,可以用dfs来做,也可以用动态规划,但明显dfs性能不如dp。
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.
动态规划:
用dis[x][y]表示到达(x,y)的最短距离,因为只能向右和向下,所以dis(x,y) = dis[x-1][y]和dis[x][y-1]的较小值加上A(x,y)即可。最终所求转化为dis(m-1,n-1)
class Solution { public: int minPathSum(vector<vector<int> > &grid) { int mm=grid.size(); int nn=grid[0].size(); vector<vector<int> > dis; vector<int> vec; vec.push_back(grid[0][0]); for (int i=1;i<nn;i++){ vec.push_back(vec[i-1]+grid[0][i]); } dis.push_back(vec); for (int i=1;i<mm;i++){ vec.clear(); for (int j=0;j<nn;j++){ if (j==0) vec.push_back(dis[i-1][j]+grid[i][j]); else vec.push_back(min(dis[i-1][j],vec[j-1])+grid[i][j]); } dis.push_back(vec); } return dis[mm-1][nn-1]; } };
除此,还应注意很多细节,几次提交都不能通过就是因为没处理好vec和dis的关系。好久没用c++做题,vector还没有push值的时候是不能通过直接访问下标进行赋值的,这一点需要特别注意。
时间: 2024-10-19 22:00:35