495. Teemo Attacking

1. 原题

In LOL 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.

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.

2. 分析

• 提莫的普攻会使艾希中毒，给出提莫的攻击的时间点， 求出艾希中毒是时间。
• 这题默认提莫只有在给定的时间点才会攻击， 所以在遍历的过程中需要考虑的一点是， 如果在前一个中毒时间点开始，到现在攻击的时候， 如果中毒状态还在，那么中毒的开始时间需要更改到当前时间，而且中毒时长增加这个间隔。

3.代码

class Solution {
public:
    int findPoisonedDuration(vector<int>& timeSeries, int duration) {
        if (timeSeries.size() == 0) return 0;

        int sum = 0;
        int begin = timeSeries[0];
        int end = begin + duration;
        sum += duration;

        for (int i = 1; i < timeSeries.size(); i++){
            if (timeSeries[i] < end){
                sum += timeSeries[i] - begin;
            }
            else
                sum += duration;

            begin = timeSeries[i];
            end = begin + duration;

        }
        return sum;
    }
};