526互联

[LeetCode] 1921. Eliminate Maximum Number of Monsters

发布时间 2023-09-03 01:03:33作者: CNoodle

You are playing a video game where you are defending your city from a group of n monsters. You are given a 0-indexed integer array dist of size n, where dist[i] is the initial distance in kilometers of the ith monster from the city.

The monsters walk toward the city at a constant speed. The speed of each monster is given to you in an integer array speed of size n, where speed[i] is the speed of the ith monster in kilometers per minute.

You have a weapon that, once fully charged, can eliminate a single monster. However, the weapon takes one minute to charge.The weapon is fully charged at the very start.

You lose when any monster reaches your city. If a monster reaches the city at the exact moment the weapon is fully charged, it counts as a loss, and the game ends before you can use your weapon.

Return the maximum number of monsters that you can eliminate before you lose, or n if you can eliminate all the monsters before they reach the city.

Example 1:

Input: dist = [1,3,4], speed = [1,1,1]
Output: 3
Explanation:
In the beginning, the distances of the monsters are [1,3,4]. You eliminate the first monster.
After a minute, the distances of the monsters are [X,2,3]. You eliminate the second monster.
After a minute, the distances of the monsters are [X,X,2]. You eliminate the thrid monster.
All 3 monsters can be eliminated.

Example 2:

Input: dist = [1,1,2,3], speed = [1,1,1,1]
Output: 1
Explanation:
In the beginning, the distances of the monsters are [1,1,2,3]. You eliminate the first monster.
After a minute, the distances of the monsters are [X,0,1,2], so you lose.
You can only eliminate 1 monster.

Example 3:

Input: dist = [3,2,4], speed = [5,3,2]
Output: 1
Explanation:
In the beginning, the distances of the monsters are [3,2,4]. You eliminate the first monster.
After a minute, the distances of the monsters are [X,0,2], so you lose.
You can only eliminate 1 monster.

Constraints:

  • n == dist.length == speed.length
  • 1 <= n <= 105
  • 1 <= dist[i], speed[i] <= 105

消灭怪物的最大数量。

你正在玩一款电子游戏，在游戏中你需要保护城市免受怪物侵袭。给你一个 下标从 0 开始 且长度为 n 的整数数组 dist ，其中 dist[i] 是第 i 个怪物与城市的 初始距离（单位：米）。

怪物以 恒定 的速度走向城市。给你一个长度为 n 的整数数组 speed 表示每个怪物的速度，其中 speed[i] 是第 i 个怪物的速度（单位：米/分）。

怪物从 第 0 分钟 时开始移动。你有一把武器，并可以 选择 在每一分钟的开始时使用，包括第 0 分钟。但是你无法在一分钟的中间使用武器。这种武器威力惊人，一次可以消灭任一还活着的怪物。

一旦任一怪物到达城市，你就输掉了这场游戏。如果某个怪物 恰 在某一分钟开始时到达城市，这会被视为 输掉 游戏，在你可以使用武器之前，游戏就会结束。

返回在你输掉游戏前可以消灭的怪物的 最大 数量。如果你可以在所有怪物到达城市前将它们全部消灭，返回  n 。

思路是贪心，具体做法是排序。因为有了怪物的距离和速度，那么我们可以得到一个 time 数组，记录每个怪物到达城市所需要的时间。这里注意需要用 double 型记录。time 数组创建好了之后我们将他排序，时间较早的排在前面。然后遍历 time 数组，我们用下标模拟时间的流逝，如果 time[i] > i （下标），说明我们可以在这个怪物未到达之间就消灭他；反之就不行，就可以立即 break 了。

时间O(nlogn)

空间O(1)

Java实现

 1 class Solution {
 2     public int eliminateMaximum(int[] dist, int[] speed) {
 3         int n = dist.length;
 4         double[] time = new double[n];
 5         for (int i = 0; i < n; i++) {
 6             time[i] = (double) dist[i] / speed[i];
 7         }
 8 
 9         Arrays.sort(time);
10         // System.out.println(Arrays.toString(time));
11         int count = 0;
12         for (int i = 0; i < n; i++) {
13             if ((double) i < time[i]) {
14                 count++;
15             } else {
16                 break;
17             }
18         }
19         return count == n ? n : count;
20     }
21 }

 

LeetCode 题目总结