leetcode解题报告（32）：Teemo Attacking

描述

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.

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.

Note:

You may assume the length of given time series array won't exceed 10000.

You may assume the numbers in the Teemo's attacking time series and his poisoning time duration per attacking are non-negative integers, which won't exceed 10,000,000.

分析

这道题并不难，只需要遍历这个时间序列，判断两个时间间隔是否大于等于duration的值，如果是，则将总伤害时间totalDur加上duration；否则，加上时间间隔。

为了避免数组越界，因此数组只遍历到倒数第二个元素，对于最后一个元素，由于是最后一次攻击，因此直接加上duration即可。

代码如下：

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