1176.Diet Plan Performance

1176. Diet Plan Performance

难度:Easy

A dieter consumes calories[i] calories on the i-th day. For every consecutive sequence of k days, they look at T, the total calories consumed during that sequence of k days:

If T < lower, they performed poorly on their diet and lose 1 point; If T > upper, they performed well on their diet and gain 1 point; Otherwise, they performed normally and there is no change in points. Return the total number of points the dieter has after all calories.length days.

Note that: The total points could be negative.

Example 1:
Input: calories = [1,2,3,4,5], k = 1, lower = 3, upper = 3
Output: 0
Explaination: calories[0], calories[1] < lower and calories[3], calories[4] > upper, total points = 0.
Example 2:
Input: calories = [3,2], k = 2, lower = 0, upper = 1
Output: 1
Explaination: calories[0] + calories[1] > upper, total points = 1.
Example 3:
Input: calories = [6,5,0,0], k = 2, lower = 1, upper = 5
Output: 0
Explaination: calories[0] + calories[1] > upper, lower <= calories[1] + calories[2] <= upper, calories[2] + calories[3] < lower, total points = 0.
Constraints:
1 <= k <= calories.length <= 10^5
0 <= calories[i] <= 20000
0 <= lower <= upper

连续k天的和的范围决定加分还是扣分,直接遍历一遍即可。

class Solution {
public:
int dietPlanPerformance(vector<int>& calories, int k, int lower, int upper) {
int startSum=0;
int points=0;
for( int i=0;i<k;i++)
startSum += calories[i];
if(startSum>upper)
++points;
else if(startSum < lower)
--points;
for(int i=k; i<calories.size();i++)
{
startSum = startSum - calories[i-k]+calories[i];
if(startSum>upper)
++points;
else if(startSum < lower)
--points;
}
return points;
}
};

执行用时 :44 ms, 在所有 C++ 提交中击败了40.28%的用户 内存消耗 :13.1 MB, 在所有 C++ 提交中击败了100.00%的用户