【数组】Leetcode编程题解:495. Teemo Attacking
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题目:
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.
样例:
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.
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.
本题要求计算艾希在被提莫攻击之后中毒的总时间,由样例可知,如果提莫攻击时艾希未处于中毒状态,则进入中毒状态;如果提莫攻击时艾希处于中毒状态,则重置中毒时间。且中毒时间不累加计算。
提交的代码:
**class Solution {
public:
int findPoisonedDuration(vector& timeSeries, int duration) {
int len = timeSeries.size(), result = 0;
if(len == 0)
return 0;
for(int i = 0; i < len - 1; i++) {
if(timeSeries[i + 1] - timeSeries[i] > duration ||
timeSeries[i + 1] - timeSeries[i] == duration)
result += duration;
else
result += (timeSeries[i + 1] - timeSeries[i]);
}
result += duration;
return result;
};**
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