【LeetCode】495. Teemo Attacking(提莫大魔王)

落日映苍穹つ 2022-04-18 02:05 118阅读 0赞

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.

**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:**  
1. You may assume the length of given time series array won’t exceed 10000.  
2. 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.

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### 题解 ###

挺有意思的，LeetCode竟然出了一题关于提莫大魔王的题目，看来英雄联盟真是火遍全世界啊。  
题目大意是：  
提莫大魔王与艾希（寒冰射手）对A，提莫可以致盲，每次致盲会有一个持续时间，致盲时间不叠加（要是叠加那主Q对面不是没得玩了），在致盲的过程中如果继续致盲，那么致盲的时间将从现在开始计算。

根据这个方案，我们可以直接扫描一次数组，计算两次间隔时间差，如果这个时间差在致盲的持续范围内，那么我们就取这个时间差，如果不在我们就可以取致盲的持续时间

### 代码 ###

public static int findPoisonedDuration(int[] timeSeries, int duration) { 
          int ret = 0;
    
           for (int i = 0; i < timeSeries.length; i++) { 
    
               if (i > 0 && timeSeries[i] - timeSeries[i - 1] < duration) { 
                   ret += timeSeries[i] - timeSeries[i - 1];
               } else { 
                   ret += duration;
               }
           }
    
           return ret;
       }

--------------------

### 题解2️⃣ ###

之前的那种方式是正向的，这次来逆向的，我们先计算总的中毒持续时间，即数组位数\*持续时间，然后跟之前一样，当出现冲突，即两次中毒时间小于持续时间，我们减去即可。

### 代码 ###

public int findPoisonedDuration(int[] timeSeries, int duration) {   
        int ret= timeSeries.length * duration;  
        for(int i=1; i<timeSeries.length; i++) {   
            if(timeSeries[i] - timeSeries[i-1] < duration) {   
                result -= duration - (timeSeries[i] - timeSeries[i-1]);  
            }  
        }  
        return ret;  
    }