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].
思路:把数组里每 K 个连续
class Solution { public double findMaxAverage(int[] nums, int k) { double sum = 0; for (int i = 0; i < k; i++) sum += nums[i]; double max = sum; for (int i = 0; i < nums.length - k; i++){ sum += nums[i+k] - nums[i]; max = Math.max(max, sum); } return max / k; } }
元素看作一个整体,然后一步一步地移动,比较这 K 个元素合与当前最大合 max 的大小。