求两个数组中字符串长度之差绝对值最大代码,Java代码库,德仔网

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【Java】求两个数组中字符串长度之差绝对值最大

作者:泓锦 / 发布于2016/2/22/ 860

给定两个字符串数组array1和array2，每一个字符串都是由a-z字母组成的。给定一个字符串x是字符串数组array1中的一个字符串，给定一个字符串y是字符串数组array2中的一个字符串，找出max(abs(length(x) ? length(y))) 如果 array1或者array2任何一个为空则返回-1。重点：其实这个问题是codewars上的一道题。主要是看了半天英文没理解。题很简单。备注：主要是求数组一和数组二中的字符串长度相减绝对值最大是什么？ array1中长度最长的字符长度串减去array2数组中长度最短的字符串的长度的绝对值和 array1中长度最短的字符长度串减去array2数组中长度最长的字符串的长度的绝对值取其中最大的那个返回。

package codewars;
 
import java.util.HashMap;
import java.util.Map;
/**
 * 问题描述
 * 
 * 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
 
 * @author pkx
 *
 */
public class MaxDiffLength {
    //人家的思路一
     public static int mxdiflg_one( String[] a1, String[] a2 ) {
            int max = -1;
            int diff = 0;
            if ( a1.length > 0 && a2.length > 0 ) {
                for (int i = 0; i < a1.length; i++) {
                    for (int j = 0; j < a2.length; j++) {
                        max =  Math.max( Math.abs( a1[i].length() - a2[j].length() ), max );
                    }
                }
            }
            return max;
        }
     //人家思路二
       public static int mxdiflg_two(String[] a1, String[] a2) {
             
            if(a1.length == 0 || a2.length == 0) return -1;
             
            int maxLength = 0;
             
            for(String a1String: a1) {
               
              for(String a2String: a2) {
               
                int difference = Math.abs(a1String.length() - a2String.length());
                if(difference > maxLength) maxLength = difference;
                 
              }
               
            }
             
            return maxLength;
           
          }
    //我的思路，提交后才能看到其他人的代码思路
    public static int mxdiflg(String a1[],String a2[]){
         
        if(a1==null ||a2==null){
            return -1;
        }
        Map<String,Integer> array1=maxAndMin(a1);
        Map<String,Integer> array2=maxAndMin(a2);
        int maxSizeOne=Math.abs(array1.get("maxSize")-array2.get("minSize"));
        int maxSieTwo=Math.abs(array1.get("minSize")-array2.get("maxSize"));
        if(maxSizeOne>maxSieTwo){
            return  maxSizeOne;
        }else{
            return maxSieTwo;
        }
         
    }
     
    public static Map<String,Integer>  maxAndMin(String str[]){
        int firstMaxStrIndex=0;
        int firstMinStrIndex=0;
         
        //字符串数组一中字符串最大长度和最小长度
        int maxStrSize=str[firstMaxStrIndex].length();
        int minStrSize=str[firstMinStrIndex].length();
        //字符串数组一
        for(int i=1;i<str.length;i++){
            //最大值
            if(maxStrSize<str[i].length()){
                maxStrSize =str[i].length();
                firstMaxStrIndex=i;
            }
            //最小值
            if(minStrSize>str[i].length()){
                firstMinStrIndex=i;
                minStrSize=str[i].length();
            }
        }
        Map<String,Integer> map = new HashMap<String,Integer>();
        map.put("maxSize", maxStrSize);
        map.put("minSize", minStrSize);
        return map;
    }
}
 
 
 
package codewars;
 
import static org.junit.Assert.*;
import org.junit.Test;
 
public class MaxDiffLengthTest {
     
    @Test
    public void test() {
        System.out.println("mxdiflg Fixed Tests");
        String[] s1 = new String[]{"hoqq", "bbllkw", "oox", "ejjuyyy", "plmiis", "xxxzgpsssa", "xxwwkktt", "znnnnfqknaz", "qqquuhii", "dvvvwz"};
        String[] s2 = new String[]{"cccooommaaqqoxii", "gggqaffhhh", "tttoowwwmmww"};
        assertEquals(13, MaxDiffLength.mxdiflg(s1, s2)); // 13
    }
}

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