Maximum Length Difference

最新推荐文章于 2022-09-05 19:59:32 发布

Z-x-H

最新推荐文章于 2022-09-05 19:59:32 发布

阅读量624

点赞数

CC 4.0 BY-SA版权

分类专栏： CodeWars Training 文章标签： abs max string

本文链接：https://blog.youkuaiyun.com/sinat_35158586/article/details/54783040

CodeWars Training 专栏收录该内容

10 篇文章

订阅专栏

本文介绍了一种计算两个字符串数组中任意两个字符串长度差绝对值最大值的方法，并提供了一个具体的Java实现示例。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Description:

You are given two arrays a1 and a2 of strings. Each string is composed with letters from a to z. Let x be any string in the first array and y be any string in the second array.

Find max(abs(length(x) − length(y)))

If a1 or a2 are empty return -1 in each language except in Haskell where you will return Nothing.

Example:

s1 = ["hoqq", "bbllkw", "oox", "ejjuyyy", "plmiis", "xxxzgpsssa", "xxwwkktt", "znnnnfqknaz", "qqquuhii", "dvvvwz"]
s2 = ["cccooommaaqqoxii", "gggqaffhhh", "tttoowwwmmww"]
mxdiflg(s1, s2) --> 13

即求出两个字符串数组 a1 跟 a2 中的任意两个字符串 x 跟 y 的长度差绝对值的最大值

My Solution:

class MaxDiffLength {

    public static int mxdiflg(String[] a1, String[] a2) {
      int max = -1;
      int difference = 0;    

      if (a1.length > 0 && a2.length >0) {
        for (int i = 0; i < a1.length; i++) {
            for (int j = 0; j < a2.length; j++) {
              difference = Math.abs(a1[i].length() - a2[j].length());
              max = Math.max(difference, max);
            }
        } 
      }
      return max;
    }

}