Lintcode 617. Maximum Average Subarray II (Medium) (Python)

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于 2018-09-12 16:20:07 发布

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分类专栏： Lintcode刷题文章标签： Lintcode Python

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本文介绍了一种寻找给定数组中长度大于等于k的最大平均子数组的方法。通过使用二分查找算法并结合滑动窗口技术，可以高效地解决这一问题。

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Maximum Average Subarray II

Description:

Given an array with positive and negative numbers, find the maximum average subarray which length should be greater or equal to given length k.

Example
Given nums = [1, 12, -5, -6, 50, 3], k = 3

Return 15.667 // (-6 + 50 + 3) / 3 = 15.667

Notice
It’s guaranteed that the size of the array is greater or equal to k.

Code:

class Solution:
    """
    @param nums: an array with positive and negative numbers
    @param k: an integer
    @return: the maximum average
    """
    def maxAverage(self, nums, k):
        # write your code here
        if not nums:
            return 0
        left, right = min(nums), max(nums)
        while right-left>0.00001:
            mid = (right+left)/2
            if self.check(nums, k, mid):
                left = mid
            else:
                right = mid
        return left

    def check(self, nums, k, mid):
        nSum = [0 for i in range(len(nums))]
        nSum[0] = nums[0]-mid

        for i in range(1, len(nums)):
            nSum[i] = nSum[i-1]+nums[i]-mid
        minSum = 0

        for i in range(k-1, len(nums)):
            if nSum[i]-minSum>=0:
                return True
            minSum = min(minSum, nSum[i-k+1])
        return False