将特殊字符保留在索引java中

Question

我有一个代码，可以使单词逐字颠倒，例如，输入是"stack overflow“，输出将是”kcat wolfrevo“，但问题是当我输入"stack，overflow”时。输出将是"，kcat .wolfrevo“而不是”kcat，wolfrevo“。

有没有办法让这个特殊字符留在他的索引中？我只知道java编程的基础知识。

我的代码：

    Scanner input = new Scanner(System.in);
    System.out.print("Enter a Paragraph: ");
    String str = input.nextLine();

    String[] Words = new String[countWords(str)];
    int temp = str.length()-1;

    for (int j = 0; j < countWords(str); j++) {

        String reverse = "";
        loop:
        for (int i=temp;i>=0;i--) {
            if (str.charAt(i) == ' ') {
                temp=temp-1;
                break;
            }
            temp=temp-1;

            reverse = reverse + str.charAt(i);
        }
        Words[j]=reverse;

    }
    for (int i = countWords(str)-1;i>=0;i--) {
        System.out.print(Words[i]+" ");
    }

}

protected static int countWords(String str) {
    int count = 0;
    if (!(" ".equals(str.substring(0, 1))) || !(" ".equals(str.substring(str.length() - 1)))) {
        for (int i = 0; i < str.length(); i++) {
            if (str.charAt(i) == ' ') {
                count++;
            }
        }
        count = count + 1;
    }
    return count;
}

Answer 1

我已经写了一个小的Python草图，它可以做你想要的。我知道它不是Java，但它的功能是基本的，应该很容易翻译成java。我添加了一些评论来指导你：

def split(string, seperators):  # parameters dont have types in Python. Imagine (String string, String separators) here
    words = []

    while True:
        j = len(string)
        if j == 0:
            break
        for sc in seperators:  # iterate over all chars in seperators string
            try:
                i = string.index(sc)
                if i < j:
                    j = i
            except ValueError:  # This is like try-catch. ValueError is thrown when no value is found. Java returns -1 when str.indexOf(char) isn't found
                continue

        nextWord = string[:j]  # this means substring(0, j)
        if nextWord:
            words.append(nextWord)
        words.append(string[j:j+1])  # this means substring(j, j+1)
        string = string[j+1:]  # this means substring(j+1)
    return words


def reverse(string):  # Imagine (String string) here
    words = split(string, "., ")
    newWords = []  # eqivalent to: new ArrayList<String>();
    for w in words:
        if len(w) > 1:
            newWords.append(w[::-1])  # w[::-1] reverses a string in python
        else:
            newWords.append(w)  # append is like ArrayList.add(e)
    return "".join(newWords)  # this joins a list to a string. In java: String.join("", list);

print(reverse("stack, overflow."))
# output: kcats, wolfrevo.

Answer 2

使用正则表达式在特殊字符处拆分输入，将这些字符保留在结果数组中；在上面的示例中，在空格char、逗号和点处拆分。然后反转结果数组中的每个元素，并将它们连接到一个新的字符串。示例：

public static void main(String[] args) {
    String input = "stack, overflow.";
    String[] splited = splitInput(input);
    String result = "";
    for(String s : splited){
        result += reverseString(s);
    }
    System.out.println(result);
}

public static String[] splitInput(String input) {
    // [,\\. ] -> split at , . ' ' and keep the delimiters
    // this will produce for the example input [stack, ,,  , overflow, .]
    String regex = "((?<=[,\\. ])|(?=[,\\. ]))";
    return input.split(regex);
}
public static String reverseString(String str) {
    return new StringBuilder(str).reverse().toString();
}

Answer 3

也许您应该考虑在这里使用堆栈数据结构，而不是数组。堆栈有一个FILO方法--您放入的第一个项目是最后一个出来的项目。考虑一下像一堆砖头或薄煎饼这样的东西。第一个放置在底部。

Here is a little driver program to show you what I mean:
import java.io.*; 
import java.util.*; 

class Test{    

    public static void main (String[] args){ 
        // creates a stack
        Stack<Integer> word_stack = new Stack<Integer>(); 

        // grab user input
        Scanner myScan = new Scanner(System.in);
        System.out.println("Please input a phrase");

        // conver it to an array
        String userInput = myScan.nextLine();
        char[] userInputArray = userInput.toCharArray();


        // pushes everything onto the stack
        for (int i = 0; i < userInputArray.length; i++){
            word_stack.stack_push(userInputArray[i]);
        }

        // pops it off
        for (i < 0; i < userInputArray.length; i++) {
            System.out.println(userInputArray[i]);
            word_stack.stack_pop();
        } //end for

    } //end psvm
} // end class