Leetcode_1266_Minimum Time Visiting All Points
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Minimum Time Visiting All Points
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
算法思路:两个点之间移动所需的最短时间,取决于横纵坐标中相差较大的。
class Solution {
public int minTimeToVisitAllPoints(int[][] points) {
int result = 0;
for(int i = 0; i < points.length - 1; i++){
int x = Math.abs(points[i][0] - points[i + 1][0]);
int y = Math.abs(points[i][1] - points[i + 1][1]);
if(x >= y)
result += x;
else
result += y;
}
return result;
}
}