[英]How to know if a given string is substring from another string in Java
嗨,我必須計算給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“包含”應該返回真,因為 str 包含 substr (否則為假)。
我在 String 類中尋找“包含”之類的東西,但沒有找到。 我想唯一的解決方案是使用模式匹配。 如果是這種情況,哪種方法會更好(最便宜)?
謝謝!
有一個contains()
方法! 它是在 Java 1.5 中引入的。 如果您使用的是早期版本,則很容易將其替換為:
str.indexOf(substr) != -1
String str="hello world";
System.out.println(str.contains("world"));//true
System.out.println(str.contains("world1"));//false
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;
}
}
如果沒有匹配項,使用 indexOf 它將返回 -1(包含在 1.5 中添加,也許您使用的是較舊的 jdk?)請參閱JDK 1.4.2 中 String 類中的“contains(CharSequence s)”方法以獲取詳細信息
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;
}
這是您可以使用的一般方法
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);
}
if (str.indexOf(substr) >= 0) {
// do something
}
我認為有一個字符串函數可以滿足您的要求:String.indexOf(String)。
請參閱此鏈接: http : //download.oracle.com/javase/1.4.2/docs/api/java/lang/String.html#indexOf(java.lang.String )
那么,你可以編寫這個函數:
public boolean isSubstring(String super, String sub) {
return super.indexOf(sub) >= 0;
}
String.indexOf(substr) 復雜度是 O(n2).. Luixv 問了一個更便宜的解決方案.. 但據我所知,沒有比當前更好的算法了。
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;
}
這將檢查 s2 是否是 s1 的子字符串。
您可以使用 .substring(int beginIndex,int lastIndex) 來檢查這個程序。 示例代碼如下:-
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;
}
}
通過這個方法。 訪問棘手的代碼
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;
}
考慮以下代碼:
如果存在子字符串,則返回給定字符串中子字符串的起始索引
否則返回-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;
}
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 ");
}
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;
}
}
}
*在他們的任何子串中,將按子串的串的第1位的形式計數*
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;
}
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