Java中如何知道给定字符串是否是另一个字符串的子字符串

Question

你好 我必须计算给定的字符串是否是更大字符串的子字符串。 例如，

String str = "Hallo my world";
String substr = "my"

方法“contains”应该返回 true，因为 str 包含 substr（否则返回 false）。

我在 String 类中寻找类似“包含”的内容 但我没有找到。我认为唯一的解决方案是使用 模式匹配。如果是这种情况，这将是更好（最便宜）的方法 要这样做吗？

谢谢！

Answer 1

您可以使用 .substring(int beginIndex,int lastIndex) 来检查此程序。示例代码如下：-

public class Test {

    public static void main(final String[] args) {
        System.out.println("Enter the first String");
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

        try {
            String s1 = br.readLine();
            System.out.println("Enter the second String");
            String s2 = br.readLine();

            boolean result = isSubStr(s1, s2);
            if (result == true)
                System.out.println("The second String is substring of the first String");
            else
                System.out.println("The second String is not a substring of the first String");

        } catch (IOException e) {
            System.out.println("Exception Caught: " + e);
        }

    }

    public static boolean isSubStr(String st1, String s2) {

        boolean result = false;

        String tem_str = "";
        int len1 = st1.length();
        int i = 0;
        int j;

        while (i < len1) {
            j = i+1;
            while (j <=len1) {
                tem_str = st1.substring(i, j);
                if (tem_str.equalsIgnoreCase(s2)) {
                    result = true;
                    break;
                }
               j++;
            }

            i++;
        }
        return result;
    }
}

Answer 2

使用indexOf，如果没有匹配，它将返回-1（contains是在1.5中添加的，也许您使用的是较旧的jdk？）请参阅JDK 1.4.2 中 String 类中的“contains(CharSequence s)”方法了解详细信息

Answer 3

  String s = "AJAYkumarReddy";
    String sub = "kumar";
    int count = 0;
    for (int i = 0; i < s.length(); i++) {
        if (s.charAt(i) == sub.charAt(count)) {
            count++;
        } else {
            count = 0;
        }
        if (count == sub.length()) {
            System.out.println("Sub String");
            return;
        }

    }

Answer 4

有一个contains() 方法！它是在 Java 1.5 中引入的。如果您使用的是早期版本，那么很容易将其替换为：

str.indexOf(substr) != -1

Answer 5

 String str="hello world";
        System.out.println(str.contains("world"));//true
        System.out.println(str.contains("world1"));//false

• Javadoc

Answer 6

这是您可以使用的通用方法

public static boolean isSubstring(String s1, String s2) {
    if(s1.length() == s2.length()) 
        return s1.equals(s2);
    else if(s1.length() > s2.length())
        return s1.contains(s2);
    else
        return s2.contains(s1);

}

Answer 7

public static boolean isSubstring(String s1, String s2){
    if(s1.length()<s2.length()) return false;
    if(s1.length()==s2.length()) return s1.equals(s2);
    for(int i=0;i<=s1.length()-s2.length();i++){
        if(s1.charAt(i)==s2.charAt(0)){
            int matchLength=1;
            for(int j=1;j<s2.length();j++){
                if(s1.charAt(i+j)!=s2.charAt(j)){
                    break;
                }
                matchLength++;
            }
            if(matchLength==s2.length()) return true;
        }
    }
    return false;
}

这将检查 s2 是否是 s1 的子字符串。

Answer 8

我认为有一个 String 函数可以满足您的要求：String.indexOf(String)。

请参阅此链接： http://download.oracle.com/javase/1.4.2/docs/api/java/lang/String.html#indexOf(java.lang.String)

那么，你可以编写这个函数：

public boolean isSubstring(String super, String sub) {
    return super.indexOf(sub) >= 0;
}

Answer 9

String.indexOf(substr) 复杂度是 O(n2) .. Luixv 提出了一种更便宜的解决方案.. 但就 而言，我知道没有比当前算法更好的算法了。

Answer 10

    public boolean isSubString(String smallStr, String largerStr) {
    char[] larger = largerStr.toCharArray();
    char[] smaller = smallStr.toCharArray();

    int i = 0;

    for (int j = 0; j < larger.length; j++) {
        if(larger[j] == smaller[i]){
            if(i == smaller.length -1){
                //done we found that this string is substring
                return true;
            }
            i++;
            continue;
        }else{
            if(i > 0){
                //that means we encountered a duplicate character before and if string was substring 
                // it shouldn't have hit this condition..
                if(larger.length - j >= smaller.length){
                    i = 0;
                    //reset i here because there are still more characters to check for substring..
                }else{
                    //we don't have enough characters to check for substring.. so done..
                    return false;
                }

            }
        }

    }

    return false;
}

Answer 11

if (str.indexOf(substr) >= 0) {
    // do something
}

Answer 12

通过这个方法。
访问棘手的代码

public static boolean isSubString(String s, String sub) {
    int count = 0;
    for (int i = 0; i < s.length(); i++) {
        if (s.charAt(i) == sub.charAt(count)) {
            count++;
        } else {
            i-=count;
            count = 0;
        }
        if (count == sub.length()) {
            return true;
        }

    }
    return false;
}

Answer 13

public class StringIsSubString {

    public static void main(String[] args) {

        String s1 = "wel";
        String s2 = "12wlecome123";

        boolean isSubStr = isSubStr(s1,s2);
        System.out.println(isSubStr);
    }

    private static boolean isSubStr(String s1, String s2) {
        String s3 = "";
        int j = 0;

        if(s1.length() > s2.length()) {
            return false;
        } else if(s1.equals(s2)){
            return true;
        } else {
            for(int i=0; i<s1.length();i++) {
                for(; j<s2.length();j++) {
                    if(s1.charAt(i) == s2.charAt(j)) {
                        s3 = s3 + s1.charAt(i);
                        break;
                    }
                }
            }
            if(s3.equals(s1)) {
                return true;
            }
            return false;       
        }
    }
}

Answer 14

*它们中的任何子字符串都将以子字符串的第1位的形式进行计数*

int isSubstring(string s1, string s2) {
    int M = s1.length();
    int N = s2.length();

    for (int i = 0; i <= N - M; i++) {
        int j;
        for (j = 0; j < M; j++)
            if (s2[i + j] != s1[j])
                break;

        if (j == M)
            return i;
    }

    return -1;
}


int main() {
    string s1 = "kumar";
    string s2 = "abhimanyukumarroy";
    int res = isSubstring(s1, s2);
    if (res == -1)
        cout << "Not present";
    else
        cout << "Present at index " << res;
    return 0;
}

Answer 15

    String str1 = "Java8 makes Java more powerful";
    String str2 = "Java";
    char c;
    char d;
    int count=0;
    boolean match = true;
    for (int i = 0; i < str1.length(); i++) {
        c = str1.charAt(i);
        for (int j = 0; j < str2.length(); j++) {
            d = str2.charAt(j);
            if (c == d) {
                match = true;
                count++;
                if(count== str2.length()){
                    i = str1.length();
                    break;
                }
                i++;
                c = str1.charAt(i);
            } else {
                match = false;
            }   
        }
    }

    if(match == true){
        System.out.println("SubString ");
    }

Answer 16

考虑以下代码：

如果子字符串存在，则返回给定字符串中子字符串的起始索引

否则返回-1

public static int isSubstring(String str, String pattern)
{
    int str_length = str.length();
    int pattern_length = pattern.length();

    for (int i = 0; i <= str_length - pattern_length; i++)
    {
        int j;

        for (j = 0; j < pattern_length; j++)
            if (str.charAt(i + j) != pattern.charAt(j))
                break;

        if (j == pattern_length)
            return i;
    }
    return -1;
}