Approach

# 計算最優路徑，看x y哪一個比較大(多走最後一步)

Time Complexity

O(n)

Space Complexity

O(n)

Code

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;
    }
};