Given an array of n positive integers and a positive integer s, find the minimal length of a contiguous subarray of which the sum ? s. If there isn't one, return 0 instead.
For example, given the array [2,3,1,2,4,3]
and s = 7
,
the subarray [4,3]
has the minimal length under the problem constraint.
思路:
类似于滑动窗口,维护一个长为k的窗口。
double findMaxAverage(vector<int>& nums, int k) { int s = accumulate(nums.begin(), nums.begin() + k, 0), m = s; for (int i = k; i < nums.size(); i++) { s += nums[i] - nums[i - k]; m = max(m, s); } return double(m) / k; }