res +=max(abs(points[i][0] - points[i +1][0]),abs(points[i][1] - points[i +1][1])); }returnres; } }; Github 同步地址: https://github.com/grandyang/leetcode/issues/1266 参考资料: https://leetcode.com/problems/minimum-time-visiting-all-points/ https://leetcode.com/problems...
代码class Solution: def minTimeToVisitAllPoints(self, points: List[List[int]]) -> int: x, y = points[0][0], points[0][1] rat = 0 for i in points: rat += max(abs(i[0] - x), abs(i[1] - y)) x, y = i[0], i[1] return rat 分类: LeetCode 标签: Array 好文...
https://leetcode.com/problems/minimum-time-visiting-all-points/ 题目描述 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...
On a plane there arenpoints with integer coordinatespoints[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 me...
The idea is to loop all points by calculating the time spent of current point and next point. Since we can either move diagonally, horizontally or vertically once at a time, it is clear the time spent is determined by the longest distance of dx or dy, because when we can consume dx an...
1266 Minimum Time Visiting All Points 访问所有点的最小时间 Description: 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: ...
points[i].length == 2 -1000 <= points[i][0], points[i][1] <= 1000 解题思路:优先走斜线,直到满足起点的x坐标等于终点的x坐标或者起点的y坐标等于终点的y坐标;然后走直线,直至到达终点。 代码如下: classSolution(object):defminTimeToVisitAllPoints(self, points):""":type points: List[List[int...
On a plane there arenpoints with integer coordinatespoints[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 me...
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