首页 > 代码库 > Minimum Path Sum
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.
class Solution {
public:
int minPathSum(vector<vector<int> > &grid) {
if(grid.size()==0 || grid[0].size()==0) return 0;
int min[grid[0].size()];
min[0]=grid[0][0];
for(int i=1;i<grid[0].size();i++)
min[i]=min[i-1]+grid[0][i];
for(int i=1;i<grid.size();i++)
{
min[0]=min[0]+grid[i][0];
for(int j=1;j<grid[0].size();j++)
{
int sum1=min[j]+grid[i][j];
int sum2=min[j-1]+grid[i][j];
min[j]=sum1;
if(sum1>sum2) min[j]=sum2;
}
}
return min[grid[0].size()-1];
}
};
public:
int minPathSum(vector<vector<int> > &grid) {
if(grid.size()==0 || grid[0].size()==0) return 0;
int min[grid[0].size()];
min[0]=grid[0][0];
for(int i=1;i<grid[0].size();i++)
min[i]=min[i-1]+grid[0][i];
for(int i=1;i<grid.size();i++)
{
min[0]=min[0]+grid[i][0];
for(int j=1;j<grid[0].size();j++)
{
int sum1=min[j]+grid[i][j];
int sum2=min[j-1]+grid[i][j];
min[j]=sum1;
if(sum1>sum2) min[j]=sum2;
}
}
return min[grid[0].size()-1];
}
};
声明：以上内容来自用户投稿及互联网公开渠道收集整理发布，本网站不拥有所有权，未作人工编辑处理，也不承担相关法律责任，若内容有误或涉及侵权可进行投诉： 投诉/举报 工作人员会在5个工作日内联系你，一经查实，本站将立刻删除涉嫌侵权内容。