[Leetcode 495] Teemo Attacking

最新推荐文章于 2021-07-13 11:41:46 发布

yangyang074

最新推荐文章于 2021-07-13 11:41:46 发布

阅读量265

点赞数

CC 4.0 BY-SA版权

分类专栏： Leetcode-C

本文链接：https://blog.youkuaiyun.com/yangyang074/article/details/73992115

Leetcode-C 专栏收录该内容

15 篇文章

订阅专栏

本文介绍了一个简单的算法问题，即如何计算游戏角色Teemo攻击敌方Ashe时造成的总中毒时间。通过给出的示例和代码实现，我们能够了解算法的基本思路及其实现过程。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Question
In LLP world, there is a hero called Teemo and his attacking can make his enemy Ashe be in poisoned condition. Now, given the Teemo’s attacking ascending time series towards Ashe and the poisoning time duration per Teemo’s attacking, you need to output the total time that Ashe is in poisoned condition.

You may assume that Teemo attacks at the very beginning of a specific time point, and makes Ashe be in poisoned condition immediately.

Example 1:
Input: [1,4], 2
Output: 4
Explanation: At time point 1, Teemo starts attacking Ashe and makes Ashe be poisoned immediately. 
This poisoned status will last 2 seconds until the end of time point 2. 
And at time point 4, Teemo attacks Ashe again, and causes Ashe to be in poisoned status for another 2 seconds. 
So you finally need to output 4.

思路
题目相当简单，只要判断
当前时间点time[i]+持续时间-1 是否大于下一时间点time[i+1]
大于: 中毒时间为 time[i+1]-time[i];
小于：中毒时间为 duration.
代码如下：

class Solution {
public:
    int findPoisonedDuration(vector<int>& timeSeries, int duration) {
        int ret=0;
        auto len=timeSeries.size();
        for(size_t i=0; i<len; ++i){
            size_t j=i+1; 
            if(j<len){
                if(timeSeries[i]+duration<=timeSeries[j])
                    ret+=duration;
                else
                    ret+=(timeSeries[j]-timeSeries[i]);
            }
            else
                ret+=duration;   //最后一个时间点，持续时间为duration
        }
        return ret;
    }
};