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[LeetCode] 1870. Minimum Speed to Arrive on Time

发布时间 2023-07-27 01:01:24作者: CNoodle

You are given a floating-point number hour, representing the amount of time you have to reach the office. To commute to the office, you must take n trains in sequential order. You are also given an integer array dist of length n, where dist[i] describes the distance (in kilometers) of the ith train ride.

Each train can only depart at an integer hour, so you may need to wait in between each train ride.

  • For example, if the 1st train ride takes 1.5 hours, you must wait for an additional 0.5 hours before you can depart on the 2nd train ride at the 2 hour mark.

Return the minimum positive integer speed (in kilometers per hour) that all the trains must travel at for you to reach the office on time, or -1 if it is impossible to be on time.

Tests are generated such that the answer will not exceed 107 and hour will have at most two digits after the decimal point.

Example 1:

Input: dist = [1,3,2], hour = 6
Output: 1
Explanation: At speed 1:
- The first train ride takes 1/1 = 1 hour.
- Since we are already at an integer hour, we depart immediately at the 1 hour mark. The second train takes 3/1 = 3 hours.
- Since we are already at an integer hour, we depart immediately at the 4 hour mark. The third train takes 2/1 = 2 hours.
- You will arrive at exactly the 6 hour mark.

Example 2:

Input: dist = [1,3,2], hour = 2.7
Output: 3
Explanation: At speed 3:
- The first train ride takes 1/3 = 0.33333 hours.
- Since we are not at an integer hour, we wait until the 1 hour mark to depart. The second train ride takes 3/3 = 1 hour.
- Since we are already at an integer hour, we depart immediately at the 2 hour mark. The third train takes 2/3 = 0.66667 hours.
- You will arrive at the 2.66667 hour mark.

Example 3:

Input: dist = [1,3,2], hour = 1.9
Output: -1
Explanation: It is impossible because the earliest the third train can depart is at the 2 hour mark.

Constraints:

  • n == dist.length
  • 1 <= n <= 105
  • 1 <= dist[i] <= 105
  • 1 <= hour <= 109
  • There will be at most two digits after the decimal point in hour.

准时到达的列车最小时速。

给你一个浮点数 hour ，表示你到达办公室可用的总通勤时间。要到达办公室，你必须按给定次序乘坐 n 趟列车。另给你一个长度为 n 的整数数组 dist ，其中 dist[i] 表示第 i 趟列车的行驶距离（单位是千米）。

每趟列车均只能在整点发车，所以你可能需要在两趟列车之间等待一段时间。

例如，第 1 趟列车需要 1.5 小时，那你必须再等待 0.5 小时，搭乘在第 2 小时发车的第 2 趟列车。
返回能满足你准时到达办公室所要求全部列车的 最小正整数 时速（单位：千米每小时），如果无法准时到达，则返回 -1 。

生成的测试用例保证答案不超过 107 ，且 hour 的 小数点后最多存在两位数字 。

来源：力扣（LeetCode）
链接：https://leetcode.cn/problems/minimum-speed-to-arrive-on-time
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思路是二分，而且是在答案上二分。题目最后让我们求的就是最小的时速，那么我们就以时速的上界和下界作为左右边界来进行二分。

在答案上二分的经典题可以参考875题，但是这道题有一个坑，就是我们在得到某一个速度，然后去计算用时的时候，for 循环不能直接走到 n，而要走到 n - 1。因为对于 dist 数组里的最后一个元素来说，因为他无需再等下一班车，所以对于最后一趟列车的用时，就是距离 / 速度的结果，无需再进位了。

时间O(n * logn)

空间O(n)

Java实现

 1 class Solution {
 2     public int minSpeedOnTime(int[] dist, double hour) {
 3         int res = -1;
 4         int left = 1;
 5         int right = (int) Math.pow(10, 7);
 6         while (left <= right) {
 7             int mid = left + (right - left) / 2;
 8             double time = helper(dist, mid);
 9             if (time <= hour) {
10                 res = mid;
11                 right = mid - 1;
12             } else {
13                 left = mid + 1;
14             }
15         }
16         return res;
17     }
18 
19     private double helper(int[] dist, int speed) {
20         double res = 0.0;
21         int n = dist.length;
22         for (int i = 0; i < n - 1; i++) {
23             res += Math.ceil((double) dist[i] / speed);
24         }
25         res += (double) dist[n - 1] / speed;
26         return res;
27     }
28 }

 

LeetCode 题目总结