Leetcode 495. 提莫攻击

原题链接：Leetcode 495. Teemo Attacking

Our hero Teemo is attacking an enemy Ashe with poison attacks! When Teemo attacks Ashe, Ashe gets poisoned for a exactly duration seconds. More formally, an attack at second t will mean Ashe is poisoned during the inclusive time interval [t, t + duration - 1]. If Teemo attacks again before the poison effect ends, the timer for it is reset, and the poison effect will end duration seconds after the new attack.

You are given a non-decreasing integer array timeSeries, where timeSeries[i] denotes that Teemo attacks Ashe at second timeSeries[i], and an integer duration.

Return the total number of seconds that Ashe is poisoned.

Example 1:

Input: timeSeries = [1,4], duration = 2
Output: 4
Explanation: Teemo’s attacks on Ashe go as follows:

• At second 1, Teemo attacks, and Ashe is poisoned for seconds 1 and 2.
• At second 4, Teemo attacks, and Ashe is poisoned for seconds 4 and 5.
  Ashe is poisoned for seconds 1, 2, 4, and 5, which is 4 seconds in total.

Example 2:

Input: timeSeries = [1,2], duration = 2
Output: 3
Explanation: Teemo’s attacks on Ashe go as follows:

• At second 1, Teemo attacks, and Ashe is poisoned for seconds 1 and 2.
• At second 2 however, Teemo attacks again and resets the poison timer. Ashe is poisoned for seconds 2 and 3.
  Ashe is poisoned for seconds 1, 2, and 3, which is 3 seconds in total.

Constraints:

• 1 <= timeSeries.length <= 10⁴
• 0 <= timeSeries[i], duration <= 10⁷
• timeSeries is sorted in non-decreasing order.

方法一：遍历

思路：

timeSeries数组已经保证了是非递减排序

C++代码：

class Solution {
   
public:
    int findPoisonedDuration(vector<int>& timeSeries, int duration) {
   
        int ans = 0, last = -1;  // last为上一次攻击的结束点
        for(int i = 0; i < timeSeries.size(); i++ ){
   
            int cur = timeSeries[i] + duration - 1; // 本次攻击持续时间
            if(last < timeSeries[i])
                ans += duration;
            else
                ans += cur - last;
            last = cur;
        }
        return ans;
    }
};