【LeetCode第 164 场周赛】回顾5271. Minimum Time Visiting All Points

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本文链接：https://blog.csdn.net/shiliang97/article/details/103223716

On a plane there are n points with integer coordinates points[i] = [xi, yi]. Your task is to find the minimum time in seconds to visit all points.

You can move according to the next rules:

In one second always you can either move vertically, horizontally by one unit or diagonally (it means to move one unit vertically and one unit horizontally in one second). You have to visit the points in the same order as they appear in the array.

Example 1:

Input: points = [[1,1],[3,4],[-1,0]] Output: 7 Explanation: One optimal path is [1,1] -> [2,2] -> [3,3] -> [3,4] -> [2,3] -> [1,2] -> [0,1] -> [-1,0] Time from [1,1] to [3,4] = 3 seconds Time from [3,4] to [-1,0] = 4 seconds Total time = 7 seconds Example 2:

Input: points = [[3,2],[-2,2]] Output: 5

Constraints:

points.length == n 1 <= n <= 100 points[i].length == 2 -1000 <= points[i][0], points[i][1] <= 1000

来源：力扣（LeetCode） 链接：https://leetcode-cn.com/problems/minimum-time-visiting-all-points 著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

class Solution {
public:
    int minTimeToVisitAllPoints(vector<vector<int>>& points) {
        int sum=0;
        for(int i=1;i<points.size();i++){
            int x=abs(points[i][0]-points[i-1][0]);
            int y=abs(points[i][1]-points[i-1][1]);
            sum+=max(x,y);
        }return sum;
    }
};