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English Version

题目描述

给定 n 个整数,找出平均数最大且长度为 k 的连续子数组,并输出该最大平均数。

 

示例:

输入:[1,12,-5,-6,50,3], k = 4
输出:12.75
解释:最大平均数 (12-5-6+50)/4 = 51/4 = 12.75

 

提示:

  • 1 <= k <= n <= 30,000。
  • 所给数据范围 [-10,000,10,000]。

解法

滑动窗口。

Python3

class Solution:
    def findMaxAverage(self, nums: List[int], k: int) -> float:
        s = sum(nums[:k])
        ans = s
        for i in range(k, len(nums)):
            s += (nums[i] - nums[i - k])
            ans = max(ans, s)
        return ans / k

Java

class Solution {
    public double findMaxAverage(int[] nums, int k) {
        int s = 0;
        for (int i = 0; i < k; ++i) {
            s += nums[i];
        }
        int ans = s;
        for (int i = k; i < nums.length; ++i) {
            s += (nums[i] - nums[i - k]);
            ans = Math.max(ans, s);
        }
        return ans * 1.0 / k;
    }
}

TypeScript

function findMaxAverage(nums: number[], k: number): number {
    let n = nums.length;
    let ans = 0;
    let sum = 0;
    // 前k
    for (let i = 0; i < k; i++) {
        sum += nums[i];
    }
    ans = sum;
    for (let i = k; i < n; i++) {
        sum += nums[i] - nums[i - k];
        ans = Math.max(ans, sum);
    }
    return ans / k;
}

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