【LeetCode】495.Teemo Attacking解题报告

【LeetCode】495.Teemo Attacking解题报告

tags: Array

题目地址：https://leetcode.com/problems/teemo-attacking/#/description
题目描述：

  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.

Example1：

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.

Example2:

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.

Solutions:

正解：

public class Solution {    public int findPoisonedDuration(int[] timeSeries, int duration) {        if (timeSeries == null || timeSeries.length == 0 || duration == 0) return 0;        int total=0;        for(int i=0;i<timeSeries.length-1;i++){            total+= timeSeries[i+1]<timeSeries[i]+duration?timeSeries[i+1]-timeSeries[i]:duration;        }        return total+duration;    }}

  其实思想很简单，就是一个时间段占位的问题，数组中各个点把时间段分开了。解题时发现其中最后一个点前面循环未处理，后面只需要在结果加上duration即可。

Date：2017年6月3日