1266. 访问所有点的最小时间

1266. 访问所有点的最小时间

• 自己画出几个例子就能发现规律，当 \(abs(x1-x2)=abs(y1-y2)\) 的长度相等时，那么最小的距离就是 \(abs(x1-x2)\)，

  否则就是两边那个较大的。

class Solution {
public:
    int minTimeToVisitAllPoints(vector<vector<int>>& points) {
        if(points.size()==1)
            return 0;
        int ans=0;
        //cout<<floor(4.9)<<endl;
        for(int i=0;i<points.size()-1;i++)
            {
                //double cnt=sqrt((points[i][0]-points[i+1][0])*(points[i][0]-points[i+1][0])+(points[i][1]-points[i+1][1])*(points[i][1]-points[i+1][1]));
               // cout<<"cnt="<<cnt<<endl;
                if(abs(points[i][0]-points[i+1][0])==abs(points[i][1]-points[i+1][1]))
                    {
                        ans+=abs(points[i][0]-points[i+1][0]);
                        //cout<<"haha"<<endl;cout<<abs(points[i][0]-points[i+1][0])<<endl;
                    }
                else
                    ans+=max(abs(points[i][0]-points[i+1][0]),abs(points[i][1]-points[i+1][1]));
            }
            //cout<<"ans="<<ans<<endl;
        return ans;

    }
};