java - 如何在不使用 Array.sort :Java 的情况下按字母顺序对数组进行排序

Question

我想在 Java 中按字母顺序对数组的值进行排序。我无法再次遍历数组并获取输出。我的意图是: 遍历单词数组， 找到最大的字符串(按字典顺序)， 找到后，将单词插入到 sortedWords 数组的末尾， 将 sortedArray 的索引向左移动一位， 通过删除已经找到的单词来减少原始数组， 再次循环查找下一个单词....

感谢您的帮助。

以下是我到目前为止所做的。

公共(public)课主要{

public static void main(String[] args) {
    String[] words = {"bob","alice","keith","zoe","sarah","gary"};
    String[] sortedWords = new String[words.length];

    // Copy of the original array
    for(int i = 0; i < sortedWords.length;i++){
        sortedWords[i]= words[i];
    }


    for (int i = 0; i < words.length - 1; i++) {
    int currentSize = words.length;
    int position = 0;
    boolean found = false;
    while (position < currentSize && !found) {
        if (words[i].equals(largestAlphabetically(words))) {
            found = true;

        } else {
            position++;
        }
    }

        if(found){
            insertAtEnd(words,largestAlphabetically(words));
            shiftLeft(sortedWords,i);
            shorterArray(words);

        }

    }
    for(int i = 0;i < sortedWords.length;i++){
        System.out.println(sortedWords[i]);
    }

}


/**
 * This method inserts the largest string lexicographically at the end of the array
 * @param words
 * @param wordToInsert
 * @return an array with string at the end
 */

public static String [] insertAtEnd(String[] words, String wordToInsert) {

    for (int i = 0; i < words.length; i++) {
        int currentSize = words.length - 1;
        wordToInsert = largestAlphabetically(words);
        if (currentSize < words.length) {
            currentSize++;
            words[currentSize - 1] = wordToInsert;
        }
    }
    return words;
}

/**
 * This method determines the largest string in an array
 * @param words
 * @return largest string lexicographically
 */
public static String largestAlphabetically(String[] words) {
    String searchedValue = words[0];
    for (int i = 0; i < words.length; i++) {

        for (int j = 0; j < words.length; j++) {
            if (words[i].compareToIgnoreCase(words[j]) < 0) {
                searchedValue = words[j];
            }
        }
    }
    return searchedValue;
}

/**
 * To shift the array index to the left
 * @param dest
 * @param from
 */

public static void shiftLeft(String[] dest, int from) {
    for (int i = from + 1; i < dest.length; i++) {
        dest[i - 1] = dest[i];
    }
    dest[dest.length - 1] = dest[0];
}

/**
 * Remove the largest word from a string while maintaining the order of the array
 * @param words
 * @return return a shorter array
 */
public static String [] shorterArray(String[] words) {
    String [] shorterArray = new String[words.length];
    int currentSize = words.length;
    String searchedValue = largestAlphabetically(words);
    int position = 0;
    boolean found = false;
    while (position < currentSize && !found) {
        if (words[position] == searchedValue) {
            found = true;

        } else {
            position++;
        }
    }
    if (found) {
        for (int i = position + 1; i < currentSize; i++) {
            words[i - 1] = words[i];

        }
        currentSize--;
        shorterArray = words;
    }

    return shorterArray;

}

Answer 1

简单的实现可以是这样的:

String[] words = {"bob","alice","keith","zoe","sarah","gary"};

boolean isSwapped = false;
do {
    isSwapped = false;
    for(int i=0;i<words.length-1;i++){
        if(words[i].compareTo(words[i+1])>0){
            String temp = words[i+1];
            words[i+1] = words[i];
            words[i] = temp;
            isSwapped = true;
        }
    }
}while((isSwapped));