LeetCode 644: Maximum Average Subarray II

最新推荐文章于 2025-02-03 15:37:10 发布

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最新推荐文章于 2025-02-03 15:37:10 发布

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原文链接：http://www.cnblogs.com/shuashuashua/p/7645855.html

本文介绍了一种寻找数组中长度为k的最大平均数值的算法实现。通过定义Solution类及findMaxAverage方法，采用二分查找策略，实现了在给定数组中找到指定长度子数组的最大平均值。同时提供了一个辅助方法checkValue用于验证中间值的有效性。

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class Solution {
    public double findMaxAverage(int[] nums, int k) {
        if (nums.length < k) return 0.0;
        double maxValue = Integer.MIN_VALUE, minValue = Integer.MAX_VALUE;
        for (int num : nums) {
            maxValue = Math.max(maxValue, num);
            minValue = Math.min(minValue, num);
        }
        double prevMid = maxValue, diff = maxValue; 
        while (diff > 0.00001) {
            double mid = (minValue + maxValue) * 0.5;
            if (checkValue(nums, mid, k)) minValue = mid;
            else maxValue = mid;
            
            diff = Math.abs(prevMid - mid);
            prevMid = mid;
        }
        return minValue;
    }
    
    private boolean checkValue(int[] nums, double mid, int k) {
        double sum = 0, prev = 0, minSum = 0;
        for (int i = 0; i < k; i++) {
            sum += nums[i] - mid;
        }
        if (sum >= 0) return true;
        for (int i = k; i < nums.length; i++) {
            sum += nums[i] - mid;
            prev += nums[i - k] - mid;
            // sum represents Sumj and prev represent Sumi. So Sumj - Sumi >= 0 should be fine.
            // Since it does not require only k windows. j - i >= k should be fine. So we just find a i' with minmum value of continue sum that let Sumj - Sumi' >= 0
 
            minSum = Math.min(prev, minSum);
            if (sum >= minSum) return true;
        }
        return false;
    }
}