[LeetCode] 495. Teemo Attacking 提莫攻击
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.
在LOL世界中，有一个名为Teemo的英雄，他的攻击可以使他的敌人Ashe处于中毒状态。 给定Teemo攻击Ashe的上升时间序列以及每个Teemo攻击的中毒持续时间，求出Ashe处于中毒状态的总时间。可以假设Teemo在特定时间点的最开始时进行攻击，并使Ashe立即处于中毒状态。
解法：题目不难，直接计算每个时间区间是否大于中毒持续时间duration，大于就累加duration。如果小于，就累加时间区间。对于最后一个攻击时间，累加duration。
Java:
public int findPoisonedDuration(int[] timeSeries, int duration) {
if (timeSeries.length == 0) return 0;
int begin = timeSeries[0], total = 0;
for (int t : timeSeries) {
total = total + (t < begin + duration ? t - begin : duration);
begin = t;
}
return total + duration;
}
Java:
public class Solution {
public int findPosisonedDuration(int[] timeSeries, int duration) {
if(timeSeries.length == 0)return 0;
if(timeSeries.length == 1)return duration;
int total = 0;
for(int i=1; i<timeSeries.length;i++)
{
total += Math.min(duration,timeSeries[i]-timeSeries[i-1]);
}
total += duration;
return total;
}
}
Java:
public class Solution {
public int findPosisonedDuration(int[] timeSeries, int duration) {
if (timeSeries == null || timeSeries.length == 0 || duration == 0) return 0;
int result = 0, start = timeSeries[0], end = timeSeries[0] + duration;
for (int i = 1; i < timeSeries.length; i++) {
if (timeSeries[i] > end) {
result += end - start;
start = timeSeries[i];
}
end = timeSeries[i] + duration;
}
result += end - start;
return result;
}
}
Python: wo
class Solution(object):
def findPoisonedDuration(self, timeSeries, duration):
"""
:type timeSeries: List[int]
:type duration: int
:rtype: int
"""
if not timeSeries:
return 0
res = 0
for i in xrange(1, len(timeSeries)):
interval = timeSeries[i] - timeSeries[i-1]
if interval >= duration:
res += duration
else:
res += interval
res += duration
return res
Python:
class Solution(object):
def findPoisonedDuration(self, timeSeries, duration):
ans = duration * len(timeSeries)
for i in range(1,len(timeSeries)):
ans -= max(0, duration - (timeSeries[i] - timeSeries[i-1]))
return ans
C++:
class Solution {
public:
int findPoisonedDuration(vector<int>& timeSeries, int duration) {
if (timeSeries.empty()) return 0;
int res = 0, n = timeSeries.size();
for (int i = 1; i < n; ++i) {
int diff = timeSeries[i] - timeSeries[i - 1];
res += (diff < duration) ? diff : duration;
}
return res + duration;
}
};
