似乎无法使其正常工作我无法按要求获得中位数

问题描述

我无法得出中位数。我想从单词中得出中位数。很难从for循环中获取价值。

public class MedianWord {
    static double medianWordLength(String words) {
        String[] parts = words.split(" ");
        int[] a;
        double median = 0;

        for (int i = 0; i < parts.length; i++) {
            a = new int[parts[i].length()];
            Arrays.sort(a);
            if (a.length % 2 == 0) {
                median = ((double) a[a.length / 2] + (double) a[a.length / 2 - 1]) / 2;
            }
            else
                median = (double) a[a.length / 2];
        }
        return median;
    }
}

解决方法

每次循环迭代都会用一个新数组覆盖a数组。您需要将问题分为两部分-首先，对parts数组进行迭代，并将其转换为长度为a的数组，然后找到该数组的中位数：

static double medianWordLength(String words) {
    String[] parts = words.split(" ");
    int[] a = new int[parts.length];
    for (int i = 0; i < parts.length; i++) {
        a[i] = parts[i].length();
    }

    Arrays.sort(a);
    if (a.length % 2 == 0) {
        return ((double) a[a.length / 2] + (double) a[a.length / 2 - 1]) / 2;
    }
    else {
        return a[a.length / 2];
    }
}

旁注：
可以使用流更优雅地将单词转换为长度排序的数组：

int[] a = Arrays.stream(words.split(" ")).mapToInt(String::length).sorted().toArray();

,

应该可以：

static Integer getMedian(String sentence) {
        String[] str = sentence.split(" ");
        Integer[] strLen = new Integer[str.length];

        for (int i = 0; i < strLen.length; i++) {
            strLen[i] = str[i].length();
        }

        return strLen.length % 2 == 0?
                (strLen[strLen.length / 2] + strLen[strLen.length / 2 - 1]) / 2 : strLen[strLen.length / 2];
    }

,

public static double medianWordLength(String str) {
    int[] arr = Arrays.stream(str.split("\\s+"))
                      .mapToInt(String::length)
                      .sorted()
                      .toArray();

    int mid = arr.length / 2;
    double median = (double)arr[mid];

    return arr.length % 2 == 0 ? (median + arr[mid - 1]) / 2 : median;
}

,

首先，您要不断对数组进行排序，然后在循环中替换数组。那是你的主要问题。

但是为了简化此过程，您只需要一个数组而不需要for循环。只需根据单词的长度对单词数组进行排序。我留了一些打印声明，以便您了解发生了什么。而且，如果您在一个或多个空格上进行分割，则在输入单词字符串时不必格外小心。

To
be
or
to
be
is
not
the
that
question
----------
be is
2.0

打印

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