[LeetCode]643. Maximum Average Subarray I 解题报告(C++)_leetcode 643. maximum average subarray i-优快云博客

本文链接：https://blog.youkuaiyun.com/qjh5606/article/details/81434380

本文介绍了解决LeetCode 643题目的两种方法：使用前缀和数组以及滑动窗口法。这两种方法都能有效地找到给定数组中长度为k的连续子数组的最大平均值。

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[LeetCode]643. Maximum Average Subarray I 解题报告(C++)

题目描述

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].

题目大意

给定数组.
找到长度为k的子串.
要求子串的均值最大.返回均值

解题思路

方法1:

p[i] 表示 nums[0] + nums[1] + nums[i];
则计算长度为k的子串和就是 p[i]-p[i-k];
遍历i从k->n-1.得到最大的和.
再除以K即可.注意返回的是 double.

代码实现:

class Solution {
public:
    double findMaxAverage(vector<int>& nums, int k) {
        int n = nums.size();
        if (n < k) {
            return 0.0;
        }
        vector<int> p(n, 0);
        p[0] = nums[0];
        // p[i]表示 num[0]+... num[i];
        for (int i = 1; i < n; i++) {
            p[i] =p[i-1]+ nums[i];
        }
        int res = p[k - 1];
        for (int i = k; i < n; i++) {
            int tmp = p[i] - p[i - k];
            res = max(res, tmp);
        }
        return res / (k*1.0);
    }
};

方法2:

维护一个滑动窗口.大小为k.
将窗口向右移动一位，即加上一个右边的数字，减去一个左边的数字
更新res

代码实现:

class Solution {
public:
    double findMaxAverage(vector<int>& nums, int k) {
        int n = nums.size();
        if (n < k) {
            return 0.0;
        }
        int sum = 0;
        for (int i = 0; i < k; ++i) {
            sum += nums[i];
        }
        int res = sum;
        for (int i = k; i < n; i++) {
            sum = sum + nums[i] - nums[i - k];
            res = max(res, sum);
        }
        return res / (k*1.0);
    }
};