Strings Mix

问题

现在有两个字符串，我们要想办法表现出他们的不同。我们只考虑字符串中的小写字符a-z。首先计算每个小写字母在两个字符串中出现的频率
s1 = “A aaaa bb c”

s2 = “& aaa bbb c d”

s1 has 4 ‘a’, 2 ‘b’, 1 ‘c’

s2 has 3 ‘a’, 3 ‘b’, 1 ‘c’, 1 ‘d’

经过比较a这个字母出现频率最高是在s1中出现4次，b是在s2中出现3次。对于那种最多的频率小于2的不考虑。
对两个字符串的不同可以表示为”1:aaaa/2:bbb”
现在给两个字符串，用以上方法表示它们的不同。

例子

s1 = “my&friend&Paul has heavy hats! &”
s2 = “my friend John has many many friends &”
mix(s1, s2) –> “2:nnnnn/1:aaaa/1:hhh/2:mmm/2:yyy/2:dd/2:ff/2:ii/2:rr/=:ee/=:ss”

s1 = “mmmmm m nnnnn y&friend&Paul has heavy hats! &”
s2 = “my frie n d Joh n has ma n y ma n y frie n ds n&”
mix(s1, s2) –> “1:mmmmmm/=:nnnnnn/1:aaaa/1:hhh/2:yyy/2:dd/2:ff/2:ii/2:rr/=:ee/=:ss”

s1=”Are the kids at home? aaaaa fffff”
s2=”Yes they are here! aaaaa fffff”
mix(s1, s2) –> “=:aaaaaa/2:eeeee/=:fffff/1:tt/2:rr/=:hh”

我的代码

package codewars;

import java.util.*;
import java.util.function.Function;
import java.util.stream.Collectors;
import java.util.stream.Stream;

public class Mixing {
    public static String mix(String s1, String s2) {
    //统计字符串m1的各个字母频率
        Map<String, Long> m1 = Stream.of(s1.replaceAll("[^a-z]","").split("")).collect(Collectors.groupingBy(Function.identity(), Collectors.counting()));
        //统计字符串m2的各个字母频率
        Map<String, Long> m2 = Stream.of(s2.replaceAll("[^a-z]","").split("")).collect(Collectors.groupingBy(Function.identity(), Collectors.counting()));
        Set<String> set1 = m1.keySet();
        Set<String> set2 = m2.keySet();
        Set<String> set3 = new HashSet<>();
        set3.addAll(set1);
        set3.addAll(set2);
        List<String> result = set3.stream().map(e -> {
            StringBuilder sb = new StringBuilder();
            if (m1.get(e) != null && m2.get(e) != null && m1.get(e) > 1 && m1.get(e) > m2.get(e)) {
                for (int i = 0; i < m1.get(e); i++) {
                    sb.append(e);
                }
                return "1:"+sb.toString();
            } else if (m1.get(e) != null && m2.get(e) != null && m1.get(e) > 1 && m1.get(e) == m2.get(e)) {
                for (int i = 0; i < m1.get(e); i++) {
                    sb.append(e);
                }
                return "=:"+sb.toString();
            } else if (m1.get(e) != null && m2.get(e) != null && m2.get(e) > 1 && m2.get(e) > m1.get(e)) {
                for (int i = 0; i < m2.get(e); i++) {
                    sb.append(e);
                }
                return "2:"+sb.toString();
            } else if (m1.get(e) != null && m2.get(e) == null && m1.get(e) > 1) {
                for (int i = 0; i < m1.get(e); i++) {
                    sb.append(e);
                }
                return "1:"+sb.toString();
            } else if (m2.get(e) != null && m1.get(e) == null && m2.get(e) > 1) {
                for (int i = 0; i < m2.get(e); i++) {
                    sb.append(e);
                }
                return "2:"+sb.toString();
            } else {
                return "";
            }
        }).filter(e -> e.length()>0).sorted(Comparator.comparing(String::length).reversed().thenComparing(Function.identity())).collect(Collectors.toList());
        return String.join("/", result);

    }

    public static void main(String[] args) {
        System.out.println(Mixing.mix(" In many languages", " there's a pair of functions"));

    }
}

分析

这道题有点难度，思路是：

1. 将两个字符串中各个小写字母的频率统计出来，结果是Map，key为字母，value为出现次数
2. 出去两个字符串中出现的所有小写字母集合set
3. 遍历set按照问题给出的逻辑拼接出字符串列表list
4. list按照长度的降序，字符串的字典序排列
5. 用”/”join出结果