[Array]495.Teemo Attacking

最新推荐文章于 2024-12-14 07:15:00 发布

Leonardo711

最新推荐文章于 2024-12-14 07:15:00 发布

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分类专栏： Leetcode

本文链接：https://blog.youkuaiyun.com/Leo00000001/article/details/71225008

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本文通过两个示例详细解析了一种算法，该算法用于计算在特定攻击模式下目标中毒的总持续时间。输入包括一系列攻击时间和中毒效果持续的时间长度。算法通过遍历攻击时间序列并判断中毒状态的叠加情况来确定总中毒时间。

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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.

Example 2:

Input: [1,2], 2
Output: 3
Explanation: At time point 1, Teemo starts attacking Ashe and makes Ashe be poisoned. 
This poisoned status will last 2 seconds until the end of time point 2. 
However, at the beginning of time point 2, Teemo attacks Ashe again who is already in poisoned status. 
Since the poisoned status won't add up together, though the second poisoning attack will still work at time point 2, it will stop at the end of time point 3. 
So you finally need to output 3.

Solution

class Solution {
public:
    int findPoisonedDuration(vector<int>& timeSeries, int duration) {
        int size = timeSeries.size();
        int i=0, j=0, tmpSum = duration, totalSum=0;
        while (i<size){
            tmpSum += timeSeries[i];
            for (j=i+1;j<size;j++){
                if (tmpSum>=timeSeries[j]) {
                   tmpSum = timeSeries[j]+duration;
                }
                else{
                    break;
                }
            }
            totalSum += (tmpSum-timeSeries[i]);
            cout << "totalSum:" << totalSum<< endl;
            i = j;
            tmpSum=duration;
        }
        return totalSum;

    }
};