[leetcode] 643. Maximum Average Subarray I
作者:互联网
Description
Given an array consisting of n integers, find the contiguous subarray of given length k that has the maximum average value. And you need to output the maximum average value.
Example 1:
Input: [1,12,-5,-6,50,3], k = 4
Output: 12.75
Explanation: Maximum average is (12-5-6+50)/4 = 51/4 = 12.75
Note:
- 1 <= k <= n <= 30,000.
- Elements of the given array will be in the range [-10,000, 10,000].
分析
题目的意思是:求其中的连续子数组,使得子数组的平均值最大。
- 我们用对数组进行求和,num[i]为前i+1个值的和,然后暴力遍历的方法来求平均值,找出最大的那个就行了。
代码
class Solution {
public:
double findMaxAverage(vector<int>& nums, int k) {
for(int i=1;i<nums.size();i++){
nums[i]+=nums[i-1];
}
double avg=0;
int i=0;
double res=INT_MIN;
while(i+k-1<nums.size()){
if(i==0){
avg=nums[k-1]/(double)k;
}else{
avg=(nums[i+k-1]-nums[i-1])/(double)k;
}
res=max(res,avg);
i++;
}
return res;
}
};
参考文献
[LeetCode] Maximum Average Subarray I 子数组的最大平均值
标签:12.75,平均值,int,average,Maximum,643,数组,Subarray,Average 来源: https://blog.csdn.net/w5688414/article/details/87893359