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...
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 ...
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: ...
LeetCode #1266. Minimum Time Visiting All Points 题目1266. Minimum Time Visiting All Points解题方法取数组中第一个点作为起点,设置一个返回值rat做累加,每次循环时rat加上新点到旧点的x、y方向上的距离的最大值,因为在平面上无论向x、y哪个方向走,由于可以走对角线,所以一定是距离更长的一个方向上需要...
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...
points[i].length == 2 -1000 <= points[i][0], points[i][1] <= 1000 解法: 任意两点之间的 花费最小时间 = max(x坐标偏移,y坐标偏移) 而这里,偏移一定为非负整数。因此计算偏移,需用到abs求绝对值。 代码参考: 1classSolution {2public:3intminTimeToVisitAllPoints(vector<vector<int>>&points)...
Polling from the front of deque ensures that we keep visiting cells with no added costs until otherwise. Once a cell is visited, we mark it and won't visit it again. The runtime is O(M * N). classSolution {publicintminCost(int[][] grid) {intm =grid.length;intn = grid[0].len...
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 (...