leetcode 1266. 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
思路:为了计算两个点的最短时间对应的路径,我们应该尽量走对角,比如(1, 1) 到 (3, 4), 通过走对角方式(1, 1) -> (2, 2) -> (3, 3) -> (3, 4), 不能直接从(1, 1)到 (3, 4), 会先走3对角,在往垂直方向1。
针对(x1, y1) -> (x2, y2), 水平方向 |x2 - x1|, 垂直方向|y2 - y1|, 走对角 min(|x2 - x1|, |y2 - y1|), 走水平或垂直max(|x2 - x1|, |y2 - y1|) - min(|x2 - x1|, |y2 - y1|), 加起来为max(|x2 - x1|, |y2 - y1|,
根据题意,可以直接贪心思想,求出相邻两点的时间,并累加。
class Solution {
public:
int minTimeToVisitAllPoints(vector<vector<int>>& points) {
int cnt = ;
for (int i = ; i < points.size(); ++i) {
cnt += max(abs(points[i][] - points[i - ][]), abs(points[i][] - points[i - ][]));
}
return cnt;
}
};
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