[英]Printing last index of string in a string without using lastIndexOf method
注意:这个问题被要求进行学校作业。 我摔倒了,我正在接近真正的代码,只剩下几点需要处理。
我被要求编写一个接收两个字符串(s1和s2)的方法,并检查s2是否在s1区分大小写。 如果s2在s1中,则返回最后一次出现的s2的索引,否则返回-1。
所以,这是我的代码:
import java.util.*;
public class homework4 {
public static void main(String args[]) {
Scanner input = new Scanner(System.in);
System.out.println("\nEnter a choice: ");
int choice = input.nextInt();
if(choice == 1) {
System.out.println("Enter firts string: ");
String s1 = input.next();
System.out.println("Enter second string: ");
String s2 = input.next();
System.out.print(contains(s1,s2));
}
else {
//Call other methods...
}
public static int contains (String s1, String s2) {
for(int i = 0; i<s1.length(); i++) {
for(int j = 0; j<s2.length(); j++) {
char ch = s2.charAt(j);
if(s1.charAt(i) == ch) {
return i;
}
}
}
return -1;
}
但是这个方法返回s2的第一个索引,或者它只是IndexOf方法的一个副本。 s1 = aabbccbbe
和s2 = bb
输出为2
。
编辑: @eli的代码
import java.util.*;
public class homework4 {
public static void main(String args[]) {
Scanner input = new Scanner(System.in);
System.out.println("\nEnter a choice: ");
int choice = input.nextInt();
if(choice == 1) {
System.out.println("Enter firts string: ");
String s1 = input.next();
System.out.println("Enter second string: ");
String s2 = input.next();
System.out.print(contains(s1,s2));
}
else {
//Call other methods...
}
public static int contains(String s1, String s2) {
int i = s2.length()-1, j = s1.length()-1;
if(i > j)
return -1;
for(; i > -1; i--) {
for(; j >= 0; j--) {
if(s1.charAt(j) == s2.charAt(i)) {
if(i == 0)
return j;
if(j != 0)
j--;
break;
} else if(i != s2.length()) {
i = s2.length()-1;
}
}
}
return -1;
}
假设你有一个叫做sentence
的字符串: 快速的棕色狐狸跳过懒狗。 并且您想要找到最后出现的“the” ,称为token
。
sentence.length = 44
和token.length = 3
考虑这个有点java伪代码:
public static int lastIndexOf(String sentence, String token) {
//The starting index is the first possible location your token could fit
int startingIndex = sentence.length() - token.length();
//move backwards one character at a time until you reach 0
//checking for string fragment that equals your token at each iteration
for (int i = startingIndex; i >= 0; i--) {
String fragment = sentence.substring(i, i + token.length());
if (fragment.equals(token)) return i;
}
return -1;
}
编辑
以下是仅使用length和charAt()的完整应用程序:
public class HelloWorld
{
// arguments are passed using the text field below this editor
public static void main(String[] args)
{
int indexOf = lastIndexOf("The quick brown fox jumps over the lazy dog.", "the");
System.out.print(indexOf);
}
public static int lastIndexOf(String sentence, String token) {
int startingIndex = sentence.length() - token.length();
for (int i = startingIndex; i >= 0; i--) {
String fragment = substring(sentence, i, i + token.length());
if (strEquals(token, fragment)) return i;
}
return -1;
}
public static String substring(String str, int startingIndex, int endingIndex) {
int size = endingIndex - startingIndex;
char[] arr = new char[size];
for (int i = 0; i < size; i++) {
arr[i] = str.charAt(startingIndex+i);
}
return new String(arr);
}
public static boolean strEquals(String s1, String s2) {
if (s1.length() != s2.length()) return false;
for (int i = 0; i < s1.length(); i++) {
if (s1.charAt(i) == s2.charAt(i)) continue;
return false;
}
return true;
}
}
编辑2
您在阅读输入的方式上也有一个错误。 您需要使用input.readLine()
来获取整行。 input.read
在空格上中断。 除此之外,您还需要为要读取的每一行使用新的扫描仪。
编辑3
以下是整个来源:
import java.util.Scanner;
public class HelloWorld {
public static void main(String[] args)
{
Scanner input1 = new Scanner(System.in);
System.out.println("\nEnter a choice: ");
String s1="";
String s2="";
int choice = input1.nextInt();
if(choice == 1) {
Scanner input2 = new Scanner(System.in);
System.out.println("Enter first string: ");
s1 = input2.nextLine();
Scanner input3 = new Scanner(System.in);
System.out.println("Enter second string: ");
s2 = input3.nextLine();
}
int indexOf = lastIndexOf(s1, s2);
System.out.println(indexOf);
}
public static int lastIndexOf(String sentence, String token) {
int startingIndex = sentence.length() - token.length();
for (int i = startingIndex; i >= 0; i--) {
String fragment = substring(sentence, i, i + token.length());
if (strEquals(token, fragment)) return i;
}
return -1;
}
public static String substring(String str, int startingIndex, int endingIndex) {
int size = endingIndex - startingIndex;
char[] arr = new char[size];
for (int i = 0; i < size; i++) {
arr[i] = str.charAt(startingIndex+i);
}
return new String(arr);
}
public static boolean strEquals(String s1, String s2) {
if (s1.length() != s2.length()) return false;
for (int i = 0; i < s1.length(); i++) {
if (s1.charAt(i) == s2.charAt(i)) continue;
return false;
}
return true;
}
}
我认为它将归结为循环字符串字符并存储发生匹配的最后一个索引。 这里不完美,但不使用indexOf
简单示例:
public static int contains(String s1, String s2) {
if(s1.length() < s2.length())
return -1;
int lastOccurrence = -1;
for (int i = 0; i < s1.length(); ) {
if (s1.startsWith(s2, i)) {
lastOccurrence = i + s2.length() - 1;
i = lastOccurrence + 1;
}
else {
++i;
}
}
return lastOccurrence;
}
首先,关闭您完成后打开的所有资源。
input.close();
如果允许,您可以使用正则表达式:
public static int contains (String s1, String s2) {
Pattern p = Pattern.compile(s2+"(?!.*"+s2+")");
Matcher m = p.matcher(s1);
if(m.find())
return m.start();
return -1;
}
使用find()
确保至少出现一次。 由于模式可以产生1且只有1个结果,因此您可以在匹配器中请求“第一次出现的第一个索引”,使用start()
实现。
编辑好吧,我可以看到除了charAt
和length
你不能使用任何东西。 这里有一个不同的解决方案,没有正则表达式,子字符串,indexOf或者什么 - 所以:
public static int contains(String s1, String s2) {
int i = s2.length()-1, j = s1.length()-1;
if(i > j)
return -1;
for(; i > -1; i--) {
for(; j >= 0; j--) {
if(s1.charAt(j) == s2.charAt(i)) {
if(i == 0)
return j;
if(j != 0)
j--;
break;
} else if(i != s2.length()) {
i = s2.length()-1;
}
}
}
return -1;
}
我必须承认我没有彻底测试这个。
最后我已经为你做了一些小修复。 我不知道你是如何编写你在帖子中编辑的内容的。 这是一个工作样本:
import java.util.Scanner;
import java.util.regex.Matcher;
import java.util.regex.Pattern;
public class homework4 {
public static void main(String args[]) {
Scanner input = new Scanner(System.in);
System.out.println("Enter choice: ");
switch (input.nextInt()) {
// If 1 is given as input...
case 1:
// As we press "enter" after inputting 1, the newline is read by the
// scanner. We skip this newline by doing this.
input.nextLine();
System.out.println("Enter first string: ");
String s1 = input.nextLine();
System.out.println("Enter second string: ");
String s2 = input.nextLine();
System.out.println("Result: " + contains(s1, s2));
break;
// If 2 is given as input (just for the sake of the example)
case 2:
System.out.println("You chose an unimplemented choice.");
break;
// If something else is given as input...
default:
System.out.println("Nothing to do...");
break;
}
// As Scanner is considered a resource, we have to close it, now that
// we're done using it.
input.close();
}
// This is the RegEx implementation
public static int containsRegx(String s1, String s2) {
Pattern p = Pattern.compile(s2 + "(?!.*" + s2 + ")");
Matcher m = p.matcher(s1);
if (m.find())
return m.start();
return -1;
}
// This is the charAt and length only
public static int contains(String s1, String s2) {
int i = s2.length() - 1, j = s1.length() - 1;
if(i > j || i * j == 0)
return -1;
for (; i > -1; i--) {
for (; j >= 0; j--) {
if (s1.charAt(j) == s2.charAt(i)) {
if (i == 0)
return j;
if (j != 0)
j--;
break;
} else if (i != s2.length()) {
i = s2.length() - 1;
}
}
}
return -1;
}
}
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