643. Maximum Average Subarray I

 

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. 1 <= k <= n <= 30,000.
  2. Elements of the given array will be in the range [-10,000, 10,000].

 

这道题给了我们一个数组nums,还有一个数字k,让我们找长度为k且平均值最大的子数组。由于子数组必须是连续的,所以我们不能给数组排序。那么怎么办呢,在博主印象中,计算子数组之和的常用方法应该是建立累加数组,然后我们可以快速计算出任意一个长度为k的子数组,用来更新结果res,从而得到最大的那个,参见代码如下:

 

解法一:

class Solution {
public:
    double findMaxAverage(vector<int>& nums, int k) {
        int n = nums.size();
        vector<int> sums = nums;
        for (int i = 1; i < n; ++i) {
            sums[i] = sums[i - 1] + nums[i];
        }
        double mx = sums[k - 1];
        for (int i = k; i < n; ++i) {
            mx = max(mx, (double)sums[i] - sums[i - k]);
        }
        return mx / k;
    }
};

 

由于这道题子数组的长度k是确定的,所以我们其实没有必要建立整个累加数组,而是先算出前k个数字的和,然后就像维护一个滑动窗口一样,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,就等同于加上右边数字减去左边数字的差值,然后每次更新结果res即可,参见代码如下:

 

解法二:

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

 

参考资料:

https://discuss.leetcode.com/topic/96134/c-simple-sliding-window-solution

https://discuss.leetcode.com/topic/96154/java-solution-sum-of-sliding-window

 

LeetCode All in One 题目讲解汇总(持续更新中…) 


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