leetcode Day18----array.easy

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本文探讨了寻找数组中三个数的最大乘积及求连续子数组最大平均值的算法实现。提供了C语言与Python3的代码示例，包括滑动窗口法等高效解决方案。

Maximum Product of Three Numbers

Given an integer array, find three numbers whose product is maximum and output the maximum product.

Example 1:

Input: [1,2,3]
Output: 6

Example 2:

Input: [1,2,3,4]
Output: 24

Note:

The length of the given array will be in range [3,104] and all elements are in the range [-1000, 1000].
Multiplication of any three numbers in the input won’t exceed the range of 32-bit signed integer.

C语言

int maximumProduct(int* nums, int numsSize){
    if(numsSize==3) return(nums[0]*nums[1]*nums[2]);
    int i=0,sec=0,third=0,max=0,min=0,min2=0;
    while(i<numsSize){
        if(nums[i]>max){third=sec;sec=max;max=nums[i];}
        else if(nums[i]>sec){third=sec;sec=nums[i];}
        else if(nums[i]>third){third=nums[i];}
        
        if(nums[i]<min){min2=min;min=nums[i];}
        else if(nums[i]<min2){min2=nums[i];}
        i++;
    }
    if(min*min2>sec*third){return max*min*min2;}
    else{return max*sec*third;}
}

Success
Details
Runtime: 36 ms, faster than 40.78% of C online submissions for Maximum Product of Three Numbers.
Memory Usage: 8.4 MB, less than 100.00% of C online submissions for Maximum Product of Three Numbers.

python3

import heapq
class Solution:
    def maximumProduct(self, nums: List[int]) -> int:
        a=heapq.nlargest(3, nums)
        b=heapq.nsmallest(2, nums)
        return max(a[0]*a[1]*a[2], b[0]*b[1]*a[0])

Success
Details
Runtime: 64 ms, faster than 97.58% of Python3 online submissions for Maximum Product of Three Numbers.
Memory Usage: 14.1 MB, less than 5.05% of Python3 online submissions for Maximum Product of Three Numbers.

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

C语言

double findMaxAverage(int* nums, int numsSize, int k){
    int sumMax=0;
    int sum=0;
    for(int i=0;i<k;i++)
    {
        sumMax=sumMax+nums[i];
    }
    sum=sumMax;
    for(int i=k;i<numsSize;i++)
    {
        sum=sum+nums[i]-nums[i-k];
        if(sum>sumMax)
        {sumMax=sum;}
    }
    return (double)sumMax/k;
}

Success
Details
Runtime: 176 ms, faster than 15.94% of C online submissions for Maximum Average Subarray I.
Memory Usage: 11.7 MB, less than 100.00% of C online submissions for Maximum Average Subarray I.

python3

sliding window,滑动窗口算法

class Solution:
    def findMaxAverage(self, nums: List[int], k: int) -> float:
        knumSum = sum(nums[:k])
        maxavg = knumSum/k
        for i in range(1, len(nums) - k + 1):
            knumSum = knumSum - nums[i-1] + nums[k+i-1]
            avg = knumSum/k
            if avg > maxavg:
                maxavg = avg
        return maxavg

Success
Details
Runtime: 176 ms, faster than 68.28% of Python3 online submissions for Maximum Average Subarray I.
Memory Usage: 15.2 MB, less than 6.25% of Python3 online submissions for Maximum Average Subarray I.