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