# 495. Teemo Attacking

495. Teemo Attacking

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:

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.

 该题的重点是在给定的事件序列中，重复的中毒事件到底有几秒钟。在两次击打之间重复的时间为可以分为下面两种情况：
① 0：两次击打之间的时间间隔超过了中毒时间
②durantion - (duration - (v[i] - v[i-1])) :两次击打之间的时间间隔小于中毒时间
 我们先假设每次击打的中毒时间都拉满，再减去根据上述讨论得出的中毒重复时间就可以得到真正的中毒时间。

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

238. Product of Array Except Self

Given an array nums of n integers where n > 1, return an array output such that output[i] is equal to the product of all the elements of nums except nums[i].

Example:

Input: [1,2,3,4]
Output: [24,12,8,6]

Note:

Please solve it without division and in O(n).

Follow up:

Could you solve it with constant space complexity? (The output array does not count as extra space for the purpose of space complexity analysis.)

创建两个数组：

①[1, 1, 1*2, 1*2*3]
②[1, 4 ,4*3, 4*3*2]

让第一个数组乘上第二个数组的倒序

class Solution {
public:
    vector<int> productExceptSelf(vector<int>& nums) {
        vector<int> num1;
        num1.push_back(1);
        vector<int> num2;
        num2.push_back(1);
        int product1 = 1;
        int product2 = 1;
        for(int i = 1; i<nums.size(); i++)
        {
            product1 = product1 * nums[i-1];
            num1.push_back(product1);
            product2 = product2 * nums[nums.size()-i];
            num2.push_back(product2);
        }
        for(int i = 0; i<num2.size(); i++)
        {
            num1[i] = num1[i] * num2[num2.size()-1-i];
        }
        return num1;
        
    }
};