原文链接: https://leetcode-cn.com/problems/minimum-speed-to-arrive-on-time

英文原文

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.

中文题目

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 。

 

示例 1：

输入：dist = [1,3,2], hour = 6
输出：1
解释：速度为 1 时：
- 第 1 趟列车运行需要 1/1 = 1 小时。
- 由于是在整数时间到达，可以立即换乘在第 1 小时发车的列车。第 2 趟列车运行需要 3/1 = 3 小时。
- 由于是在整数时间到达，可以立即换乘在第 4 小时发车的列车。第 3 趟列车运行需要 2/1 = 2 小时。
- 你将会恰好在第 6 小时到达。

示例 2：

输入：dist = [1,3,2], hour = 2.7
输出：3
解释：速度为 3 时：
- 第 1 趟列车运行需要 1/3 = 0.33333 小时。
- 由于不是在整数时间到达，故需要等待至第 1 小时才能搭乘列车。第 2 趟列车运行需要 3/3 = 1 小时。
- 由于是在整数时间到达，可以立即换乘在第 2 小时发车的列车。第 3 趟列车运行需要 2/3 = 0.66667 小时。
- 你将会在第 2.66667 小时到达。

示例 3：

输入：dist = [1,3,2], hour = 1.9
输出：-1
解释：不可能准时到达，因为第 3 趟列车最早是在第 2 小时发车。

 

提示：

• n == dist.length
• 1 <= n <= 105
• 1 <= dist[i] <= 105
• 1 <= hour <= 109
• hours 中，小数点后最多存在两位数字

通过代码

高赞题解

解题思路

首先判断特殊情况：当所需时间向上取整后仍然小于 dist 长度，那么注定到不了，就直接返回 false。
然后进行二分搜索，范围是 [1, Integer.MAX_VALUE]。

代码

class Solution {
    public int minSpeedOnTime(int[] dist, double hour) {
        if (dist.length > Math.ceil(hour)) return -1;
        // 搜索边界
        int left = 1, right = Integer.MAX_VALUE;
        while (left < right) {
            int mid = left + (right - left) / 2;
            // 如果以 mid 速度可达，那么就尝试减小速度
            if (check(dist, hour, mid)) right = mid;
            // 否则就需要加了
            else left = mid + 1;
        }
        return left;
    }

    private boolean check(int[] dist, double hour, int speed) {
        double cnt = 0.0;
        // 对除了最后一个站点以外的时间进行向上取整累加
        for (int i = 0; i < dist.length - 1; ++i) {
            // 除法的向上取整
            cnt += (dist[i] + speed - 1) / speed;
        }
        // 加上最后一个站点所需的时间
        cnt += (double) dist[dist.length - 1] / speed;
        return cnt <= hour;
    }
}

统计信息

通过次数	提交次数	AC比率
7757	17193	45.1%

提交历史

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