2020-03-17
1.题目描述
访问所有点的最小时间
2.题解
两个点之间的最小访问时间为两个点之间横纵坐标差绝对值的最大值
3.代码
#include <iostream>
#include <vector>
#include <cmath>
using namespace std;
class Solution {
public:
int minTimeToVisitAllPoints(vector<vector<int>>& points) {
int cnt=0,l=points.size();
vector<int>t1,t2;
for (int i=1;i<l;i++){
t1=points[i-1];t2=points[i];
cnt+=max(abs(t1[0]-t2[0]),abs(t1[1]-t2[1]));
}
return cnt;
}
};
int main(){
Solution s;
vector<int>t1={3,2};
vector<int>t2={-2,2};
// vector<int>t3={-1,0};
vector<vector<int>>p;
p.push_back(t1);
p.push_back(t2);
// p.push_back(t3);
cout<<s.minTimeToVisitAllPoints(p)<<endl;
return 0;
}