Question

需要Java函数来查找两个字符串的交集。即字符串共有的字符。

示例：

String s1 = new String("Sychelless");
String s2 = new String("Sydney");

Answer 1

使用HashSet<Character>：

HashSet<Character> h1 = new HashSet<Character>(), h2 = new HashSet<Character>();
for(int i = 0; i < s1.length(); i++)                                            
{
  h1.add(s1.charAt(i));
}
for(int i = 0; i < s2.length(); i++)
{
  h2.add(s2.charAt(i));
}
h1.retainAll(h2);
Character[] res = h1.toArray(new Character[0]);

这是O(m + n)，这是渐近最优的。

Answer 2

提取字符

String.toCharArray

将它们放入一套找到交叉点

Set.retainAll

Answer 3

最基本的方法：

String wordA = "Sychelless";  
String wordB = "Sydney";  
String common = "";  

for(int i=0;i<wordA.length();i++){  
    for(int j=0;j<wordB.length();j++){  
        if(wordA.charAt(i)==wordB.charAt(j)){  
            common += wordA.charAt(i)+" ";  
            break;
        }  
    }  
}  
System.out.println("common is: "+common);

Answer 4

关于saugata的回应的更多细节（在我写这篇文章时出现）： -

public static void main(String[] args) {
    String s1 = "Seychelles";
    String s2 = "Sydney";
    Set<Character> ss1 = toSet(s1);
    ss1.retainAll(toSet(s2));
    System.out.println(ss1);
}

public static Set<Character> toSet(String s) {
    Set<Character> ss = new HashSet<Character>(s.length());
    for (char c : s.toCharArray())
        ss.add(Character.valueOf(c));
    return ss;
}

Answer 5

我认为您正在寻找的算法是problem of the longest common subsequence

Answer 6

在此处找到相同的问题，请参阅此

Implementing an efficent algorithm to find the intersection of two strings

Answer 7

优化的解决方案：

public static String twoStrings(String s1, String s2){

    HashSet<Character> stringOne =  new HashSet<Character>(), stringTwo = new HashSet<Character>();  
    int stringOneLength = s1.length();
    int stringTwoLength = s2.length();
    for(int i=0; i<stringOneLength || i<stringTwoLength; i++) {
        if(i < stringOneLength)
            stringOne.add(s1.charAt(i));
        if(i < stringTwoLength)
            stringTwo.add(s2.charAt(i));
    }
    stringOne.retainAll(stringTwo);

    return stringOne.toString();
}

Answer 8

通过番石榴，这项任务似乎更容易：

String s1 = new String("Sychelless");
String s2 = new String("Sydney");
Set<String> setA = Sets.newHashSet(Splitter.fixedLength(1).split(s1));
Set<String> setB = Sets.newHashSet(Splitter.fixedLength(1).split(s2));
Sets.intersection(setA, setB);

Answer 9

我使用过TreeSet。 TreeSet中retainAll()获取匹配的元素。

Oracle Doc：
retainAll(Collection<?> c)
仅保留此集合中包含的元素指定集合（可选操作）。

String s1 = new String("Sychelless");
String s2 = new String("Sydney");

Set<Character> firstSet = new TreeSet<Character>();
for(int i = 0; i < s1.length(); i++) {
    firstSet.add(s1.charAt(i));
}

Set<Character> anotherSet = new TreeSet<Character>();
for(int i = 0; i < s2.length(); i++) {
    anotherSet.add(s2.charAt(i));
}

firstSet.retainAll(anotherSet);
System.out.println("Matched characters are " + firstSet.toString());//print common strings

//output > Matched characters are [S, e, y]

Answer 10

s1.contains(s2) returns true;
s1.indexOf(s2) returns 0. 
s1.indexOf("foo") returns -1

对于更复杂的案例，请使用类Pattern。

Java中两个字符串的交集

10 个答案: