【Leetcode 1266】 Minimum Time Visiting All Points（C++）

题目

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

源代码

int minTimeToVisitAllPoints(vector>& points) {
	int sum = 0;
	for (int i = 0; i < points.size() - 1; i++)
	{
		int a =abs(points[i][0] - points[i + 1][0]);
		int b = abs(points[i][1] - points[i + 1][1]);
		sum += max(a,b);
	}
	return sum;
}