week17- NO.495. Teemo Attacking

题目

• Total Accepted: 12023
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• Difficulty: Medium
• Contributors:love_Fawn

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], 2Output: 4Explanation: 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], 2Output: 3Explanation: 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.

https://leetcode.com/problems/teemo-attacking/#/description

思路

题目给定一个整数数组，和一个整数duration，一个英雄会在数组中给定的时间放毒，毒持续的时间是duration，中毒伤害不叠加，希望求得英雄最终能造成多少伤害。

遍历数组中的时间点，若一个时间点和前一个时间点造成的中毒状态没有重叠，则伤害增加duration，否则减去重叠的时间。

源程序

class Solution {public:    int findPoisonedDuration(vector<int>& timeSeries, int duration) {        if(timeSeries.size() == 0 || duration == 0)            return 0;        int result = 0;        int i;        result += duration;        for(i = 1;i < timeSeries.size();i ++){            result += duration;            if(timeSeries[i - 1] + duration - timeSeries[i] > 0)                result -= (timeSeries[i - 1] + duration - timeSeries[i]);        }        return result;        }};