# DAY 2
0
Software Development

## 題目

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, calories < lower and calories, calories > upper, total points = 0.
``````

Example 2:

``````Input: calories = [3,2], k = 2, lower = 0, upper = 1
Output: 1
Explaination: calories + calories > upper, total points = 1.
``````

Example 3:

``````Input: calories = [6,5,0,0], k = 2, lower = 1, upper = 5
Output: 0
Explaination: calories + calories > upper, calories + calories < lower, total points = 0.
``````

## 解法

``````class Solution {
public int dietPlanPerformance(int[] calories, int k, int lower, int upper) {
int len = calories.length, sum = 0;
for(int i = 0; i+k-1 < len; i++){
int temp = 0;
for (int j=0; j<k; j++){
temp += calories[i+j];
}
if(temp < lower){
sum--;
}
if(temp > upper){
sum++;
}
}
return sum;
}
}
``````

LeetCode小試身手14