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].
给定一个由n个整数组成的数组,计算数组中相邻连续k个元素的最大平均值。按照题意使用了两层for循环计算结果,但是提交时显示超时。
class Solution { public: double findMaxAverage(vector<int>& nums, int k) { double res = -INT_MAX; for (int i = 0; i != nums.size() - k + 1; i++) { double sum = 0; for (int j = i; j != i + k; j++) sum += nums[j]; res = max(res, sum); } return res / k; } }; // TLE
分析原因得到2层for循环导致超时,经过思考采用一个for循环来进行计算,每次计算到k的整数和时判断该和是否为最大值,接着把最k个数中第一个减去,再在后面添加一个整数。类似于滑动一个长度为k的窗口,代码如下
class Solution { public: double findMaxAverage(vector<int>& nums, int k) { double res = INT_MIN; double sum = 0; for (int i = 0; i != nums.size(); i++) { sum += nums[i]; if (i >= k) sum -= nums[i - k]; if (i >= k - 1) res = max(res, sum); } return res / k; } }; // 206 ms