leetcode每日一题-495:提莫攻击

链接

提莫攻击

题目

分析

从第二位开始枚举,每次判断和前一次的间隔是多少,然后判断出中毒的描述,遍历结束后,因为最后一次施加的中毒状态一定可以持续完整的中毒事件,所以最后再加上duration.

代码

C++

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

        return res + duration;
    }
};

Java

class Solution {
    public int findPoisonedDuration(int[] timeSeries, int duration) {
        int ans = 0;
        int expired = 0;
        for (int i = 0; i < timeSeries.length; ++i) {
            if (timeSeries[i] >= expired) {
                ans += duration;
            } else {
                ans += timeSeries[i] + duration - expired;
            }
            expired = timeSeries[i] + duration;
        }
        return ans;
    }
}

作者：LeetCode-Solution