n 个字符串中的常用字符

Question

我正在尝试制作一个函数来打印给定 n 个字符串中常见的字符数。（注意字符可以多次使用）

我正在努力对 n 个字符串执行此操作但是我为 2 个字符串执行了此操作，而没有任何字符重复多次。

我已经发布了我的代码。

public class CommonChars {

    public static void main(String[] args) {

        String str1 = "abcd";

        String str2 = "bcde";

        StringBuffer sb = new StringBuffer();

        // get unique chars from both the strings

        str1 = uniqueChar(str1);

        str2 = uniqueChar(str2);

        int count = 0;

        int str1Len = str1.length();

        int str2Len = str2.length();

        for (int i = 0; i < str1Len; i++) {

            for (int j = 0; j < str2Len; j++) {

                // found match stop the loop

                if (str1.charAt(i) == str2.charAt(j)) {

                    count++;

                    sb.append(str1.charAt(i));

                    break;

                }

            }

        }   

        System.out.println("Common Chars Count : " + count + "\nCommon Chars :" + 

        sb.toString());

    }

    public static String uniqueChar(String inputString) {

        String outputstr="",temp="";

        for(int i=0;i<inputstr.length();i++) {

            if(temp.indexOf(inputstr.charAt(i))<0) {

                temp+=inputstr.charAt(i);

            }

        }

        System.out.println("completed");

        return temp;

    }

}

3 

abcaa

bcbd

bgc

3

他们可能有机会在一个字符串中多次出现相同的字符，并且您不应该消除这些字符而是检查否。他们在其他字符串中重复的次数。例如

 3

 abacd

 aaxyz

 aatre

输出应该是 2

如果我在java中得到解决方案会更好

Answer 1

您必须将所有Strings 转换为Setof Characters 并保留第一个中的所有内容。下面的解决方案有很多可以优化的地方，但你应该理解一般的想法。import java.util.Arrays;import java.util.Collection;import java.util.Collections;import java.util.HashSet;import java.util.List;import java.util.Set;public class Main {&nbsp; &nbsp; public static void main(String[] args) {&nbsp; &nbsp; &nbsp; &nbsp; List<String> input = Arrays.asList("jonas", "ton", "bonny");&nbsp; &nbsp; &nbsp; &nbsp; System.out.println(findCommonCharsFor(input));&nbsp; &nbsp; }&nbsp; &nbsp; public static Collection<Character> findCommonCharsFor(List<String> strings) {&nbsp; &nbsp; &nbsp; &nbsp; if (strings == null || strings.isEmpty()) {&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; return Collections.emptyList();&nbsp; &nbsp; &nbsp; &nbsp; }&nbsp; &nbsp; &nbsp; &nbsp; Set<Character> commonChars = convertStringToSetOfChars(strings.get(0));&nbsp; &nbsp; &nbsp; &nbsp; strings.stream().skip(1).forEach(s -> commonChars.retainAll(convertStringToSetOfChars(s)));&nbsp; &nbsp; &nbsp; &nbsp; return commonChars;&nbsp; &nbsp; }&nbsp; &nbsp; private static Set<Character> convertStringToSetOfChars(String string) {&nbsp; &nbsp; &nbsp; &nbsp; if (string == null || string.isEmpty()) {&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; return Collections.emptySet();&nbsp; &nbsp; &nbsp; &nbsp; }&nbsp; &nbsp; &nbsp; &nbsp; Set<Character> set = new HashSet<>(string.length() + 10);&nbsp; &nbsp; &nbsp; &nbsp; for (char c : string.toCharArray()) {&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; set.add(c);&nbsp; &nbsp; &nbsp; &nbsp; }&nbsp; &nbsp; &nbsp; &nbsp; return set;&nbsp; &nbsp; }}上面的代码打印：[n, o]

Answer 2

好吧，如果一个人去散列：public static int uniqueChars(String first, String second) {&nbsp; &nbsp; boolean[] hash = new boolean[26];&nbsp;&nbsp; &nbsp; int count = 0;&nbsp; &nbsp; //reduce first string to unique letters&nbsp; &nbsp; for (char c : first.toLowerCase().toCharArray()) {&nbsp; &nbsp; &nbsp; &nbsp; hash[c - 'a'] = true;&nbsp; &nbsp; }&nbsp; &nbsp; //reduce to unique letters in both strings&nbsp; &nbsp; for(char c : second.toLowerCase().toCharArray()){&nbsp; &nbsp; &nbsp; &nbsp; if(hash[c - 'a']){&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; count++;&nbsp;&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; hash[c - 'a'] = false;&nbsp; &nbsp; &nbsp; &nbsp; }&nbsp; &nbsp; }&nbsp; &nbsp; return count;}这是使用 bucketsort，它给出了 n+m 的复杂度，但需要 26 个桶（“散列”数组）。Imo 在复杂性方面不能做得更好，因为您需要至少查看每个字母一次，总和为 n + m。Insitu 你能得到的最好的结果是在 O(n log(n) ) 范围内的某个地方。你的方法是在 O(n²) 的联盟中的某个地方插件：如果您需要将字符作为字符串（本质上与上面相同，计数是返回的字符串的长度）：public static String uniqueChars(String first, String second) {&nbsp; &nbsp; boolean[] hash = new boolean[26];&nbsp;&nbsp; &nbsp; StringBuilder sb = new StringBuilder();&nbsp; &nbsp; for (char c : first.toLowerCase().toCharArray()) {&nbsp; &nbsp; &nbsp; &nbsp; hash[c - 'a'] = true;&nbsp; &nbsp; }&nbsp; &nbsp; for(char c : second.toLowerCase().toCharArray()){&nbsp; &nbsp; &nbsp; &nbsp; if(hash[c - 'a']){&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; sb.append(c);&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; hash[c - 'a'] = false;&nbsp; &nbsp; &nbsp; &nbsp; }&nbsp; &nbsp; }&nbsp; &nbsp; return sb.toString();}

Answer 3

获取每个字符串的字符列表：List<Character> chars1 = s1.chars()&nbsp; &nbsp; // list of chars for first string&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; .mapToObj(c -> (char) c)&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; .collect(Collectors.toList());List<Character> chars2 = s2.chars()&nbsp; &nbsp; // list of chars for second string&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; .mapToObj(c -> (char) c)&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; .collect(Collectors.toList());然后使用retainAll方法：chars1.retainAll(chars2);&nbsp; // retain in chars1 only the chars that are contained in the chars2 alsoSystem.out.println(chars1.size());如果您想获得唯一字符的数量，只需使用Collectors.toSet()而不是toList()

Answer 4

public static String getCommonCharacters(String... words) {&nbsp; &nbsp; if (words == null || words.length == 0)&nbsp; &nbsp; &nbsp; &nbsp; return "";&nbsp; &nbsp; Set<Character> unique = words[0].chars().mapToObj(ch -> (char)ch).collect(Collectors.toCollection(TreeSet::new));&nbsp; &nbsp; for (String word : words)&nbsp; &nbsp; &nbsp; &nbsp; unique.retainAll(word.chars().mapToObj(ch -> (char)ch).collect(Collectors.toSet()));&nbsp; &nbsp; return unique.stream().map(String::valueOf).collect(Collectors.joining());}另一个不创建临时变量Set并使用Character.public static String getCommonCharacters(String... words) {&nbsp; &nbsp; if (words == null || words.length == 0)&nbsp; &nbsp; &nbsp; &nbsp; return "";&nbsp; &nbsp; int[] arr = new int[26];&nbsp; &nbsp; boolean[] tmp = new boolean[26];&nbsp; &nbsp; for (String word : words) {&nbsp; &nbsp; &nbsp; &nbsp; Arrays.fill(tmp, false);&nbsp; &nbsp; &nbsp; &nbsp; for (int i = 0; i < word.length(); i++) {&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; int pos = Character.toLowerCase(word.charAt(i)) - 'a';&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; if (tmp[pos])&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; continue;&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; tmp[pos] = true;&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; arr[pos]++;&nbsp; &nbsp; &nbsp; &nbsp; }&nbsp; &nbsp; }&nbsp; &nbsp; StringBuilder buf = new StringBuilder(26);&nbsp; &nbsp; for (int i = 0; i < arr.length; i++)&nbsp; &nbsp; &nbsp; &nbsp; if (arr[i] == words.length)&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; buf.append((char)('a' + i));&nbsp; &nbsp; return buf.toString();}演示System.out.println(getCommonCharacters("abcd", "bcde"));&nbsp; // bcd