Leetcode 495. Teemo Attacking

题目

链接：https://leetcode.com/problems/teemo-attacking/

Level: Easy

Discription:
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.

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.

代码

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

思考

• 算法时间复杂度为O(N),空间复杂度为O(1)。
• 这题比较简单，只需要比较时间差和间隔之间的最小值累加即可，最后不要忘了最后一次的叠加，注意时间序列的长度可能为0。

转载于:https://www.cnblogs.com/zuotongbin/p/10214940.html