解题思路:
(1)先使用对角走完两个点坐标差绝对值较小的距离,接下来再平移
(2)按照这种走法,其实两点之间距离即为横纵坐标中相差绝对值最大的距离
int minTimeToVisitAllPoints(vector<vector<int>>& points) {
int i = 0,distance = 0;
while(i+1<points.size()) {
distance += max(abs(points[i][0]-points[i+1][0]),abs(points[i][1]-points[i+1][1]));
i++;
}
return distance;
}