LeetCode 495. Teemo Attacking-优快云博客

本文链接：https://blog.youkuaiyun.com/lmxr133/article/details/55672114

本文介绍了一种算法，用于计算角色Ashe在受到Teemo攻击后处于中毒状态的总时间。通过分析Teemo攻击的时间序列和每次攻击导致的中毒持续时间，算法能够准确计算出中毒总时间。

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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.

public class Solution {
    public int findPoisonedDuration(int[] timeSeries, int duration) {
        int result = 0;
        for(int i=0; i<timeSeries.length; i++){
            //第一个元素没有特殊情况，直接累计
            if(i == 0){
                result += duration;
            }else{
                //分两种情况处理
                //第一种情况，前一个元素+持续时间，他们二者之和覆盖了后一个元素计算持续时间
                //这时需要用该元素的持续时间——上一个元素和持续时间的和超出的部分，最后的那个1是由于覆盖的第一个元素不算，所以需要减去
                if((timeSeries[i]-timeSeries[i-1]+1) <= duration){
                    result += (duration -(duration - (timeSeries[i] - timeSeries[i-1] + 1) + 1)); 
                }else{
                    //第二种情况，前一个元素+持续时间，总是小于后面一个元素
                    result += duration;
                }
            }
            
        }
        return result;
    }
}