Computer science note/LeetCode
Maximum Average Subarray I
Leetcode problem description and solution for Maximum Average Subarray I
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You are given an integer array nums
consisting of n
elements, and an integer k
.
您将获得一个由 n
元素组成的整数数组 nums
和一个整数 k
.
Find a contiguous subarray whose length is equal to k
that has the maximum average value and return this value. Any answer with a calculation error less than 10-5
will be accepted.
找到一个长度等于具有最大平均值 k
的连续子数组,并返回此值。任何计算误差小于 10-5
的答案都将被接受。
Example 1: 示例 1:
Input: nums = [1,12,-5,-6,50,3], k = 4 Output: 12.75000 Explanation: Maximum average is (12 - 5 - 6 + 50) / 4 = 51 / 4 = 12.75
Example 2: 示例 2:
Input: nums = [5], k = 1 Output: 5.00000
Constraints: 约束:
n == nums.length
1 <= k <= n <= 105
-104 <= nums[i] <= 104
package org.leetcode;
public class MaximumAverageSubarrayI {
public double findMaxAverage(int[] nums, int k) {
int sum = 0;
for(int i = 0; i < k; i++) {
sum = sum + nums[i];
}
int i = k;
int j = 0;
double result = sum/k;
while(i < nums.length) {
sum += nums[i];
sum -= nums[j];
result = Math.max(result, sum/k);
i++;
j++;
}
return result;
}
public static void main(String[] args) {
MaximumAverageSubarrayI maximumAverageSubarrayI = new MaximumAverageSubarrayI();
int[] nums1 = {1,12,-5,-6,50,3};
int k1 = 4;
double result1 = maximumAverageSubarrayI.findMaxAverage(nums1, k1);
double expected1 = 12.75;
if(result1 == expected1) {
System.out.println("Test 1 passed");
} else {
System.out.println("Test 1 failed expected: " + expected1 + " actual: " + result1);
}
int[] nums2 = {5};
int k2 = 1;
double result2 = maximumAverageSubarrayI.findMaxAverage(nums2, k2);
double expected2 = 5;
if(result2 == expected2) {
System.out.println("Test 2 passed");
} else {
System.out.println("Test 2 failed expected: " + expected2 + " actual: " + result2);
}
}
}