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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].
题意:给定一个一维数组num,和一个长度值len,要求找出数组中长度为len的连续子数组,使其满足所有元素的和的平均值最大;
解法一:最简单的方法就是遍历所有的可能,找出所有满足长度为len的连续子数组,比较得到其最大的平均值;
Java
class Solution {
public double findMaxAverage(int[] nums, int k) {
double maxAverage = Integer.MIN_VALUE;
for(int i=0; i<nums.length; i++){
if (i+k > nums.length) {
break;
}
int sum = 0;
for(int j=i; j<i+k; j++) {
sum += nums[j];
}
maxAverage = Math.max(maxAverage, (double)sum / k);
}
return maxAverage;
}
}
解法二:在第一种解法中,我们每次都要计算k个元素的和后再求平均值;因此,我们可以定义一个数组sum,数组中的元素xi表示给定数组中从下标0开始到下标i的所有元素的和,这样,我们在求解k个元素的和时可以利用sum[i] - sum[i-k]来得到;
Java
class Solution {
public double findMaxAverage(int[] nums, int k) {
int[] sum = new int[nums.length];
sum[0] = nums[0];
for(int i=1; i<nums.length; i++) {
sum[i] = sum[i-1] + nums[i];
}
double maxAverage = sum[k-1] / (double)k;
for(int i=k; i<nums.length; i++) {
maxAverage = Math.max(maxAverage, (sum[i] - sum[i-k]) / (double)k);
}
return maxAverage;
}
}
解法三:对于第二种解法来说,我们需要额外的开销用来存储元素的和;观察下面的公式;
对于从i -> i+k来说,子数组的和为:
对于从i+1 -> i+k+1来说,子数组的和为:
我们会发现,其中有k-1个元素是重复的,因此我们计算过第一个长度为k的子数组的和sum后,那么其下一个子数组(假设其结束下标位置为j)的和应该是sum +nums[j] - nums[j-k];
Java
class Solution {
public double findMaxAverage(int[] nums, int k) {
int sum = 0;
for(int i=0; i<k; i++) {
sum += nums[i];
}
double maxAverage = sum / (double)k;
for(int i=k; i<nums.length; i++) {
sum += nums[i] - nums[i-k];
maxAverage = Math.max(maxAverage, sum / (double)k);
}
return maxAverage;
}
}