Question

注意：此问题会被要求进行学校作业。我感到沮丧我接近真正的代码，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）和的方法检查s2是否在s1情况下敏感。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的输出为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;
    }

Answer 1

假设您有一个名为sentence的字符串：快速的棕色狐狸跳过懒狗。你想找到最后一次出现的＆＃34;＆＃ 34; ，名为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;
}

修改

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()来获得完整的一行。 input.read打破空格。沿着这些方向，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; } }

Answer 2

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的简单示例：

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;
}

Answer 3

首先，当您完成时，关闭所有打开的资源。

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;
}

解释正则表达式模式here。

使用find()，您可以确保至少出现一次。由于模式可以产生1且只有1个结果，因此您可以只询问第一次出现的＆＃34;第一个索引＆＃34;在匹配器中，使用start()实现。

修改好的，我可以看到除了charAt和length之外你不能使用任何其他内容。这里有一个不同的解决方案，没有正则表达式，子字符串，indexOf或者以前的所有：

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过这个。

<强> FINAL 我为你做了一些小修理。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; } }

不使用lastIndexOf方法在字符串中打印字符串的最后一个索引

3 个答案: