[LeetCode] 1870. Minimum Speed to Arrive on Time
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 takes1.5
hours, you must wait for an additional0.5
hours before you can depart on the2nd
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 }