验证字符串列表中的自定义排序

Question

我正在尝试验证返回的列表是否已排序，但通过预定义的排序算法。如果列表未排序，那么我应该通过自定义排序对其进行排序。

我已经实现了自定义排序功能，但是我想验证检索到的列表是否根据自定义排序进行排序。

这是自定义排序的代码：

public static void customSort(final String order, String[] array){
        String[] alphabets = {"!", "@", "#", "$", "%", "^", "&", "(", ")", "*", "+", ",", "-", "/",
                "0", "1", "2", "3", "4", "5", "6", "7", "8", "9", "=",
                "a", "b", "c", "d", "e", "f", "g", "h", "i", "j", "k", "l", "m", "n", "o", "p", "q", "r", "s", "t", "u", "v", "w", "x",  "y", "z",
                "A", "B", "C", "D", "E", "F", "G", "H", "I", "J", "K", "L", "M", "N", "O", "P", "Q", "R", "S", "T", "U", "V", "W", "X", "Y", "Z",
                "[", "]", "_"
        };
        String keyword = order;
        for(int g = 0; g < alphabets.length; g++){
            String one = alphabets[g];
            if(!keyword.contains(one)){
                keyword = keyword + one;
            }
        }

        final String finalKeyword = keyword;
        Arrays.sort(array, new Comparator<String>() {

            @Override
            public int compare(String o1, String o2) {
                int pos1 = 0;
                int pos2 = 0;
                for (int i = 0; i < Math.min(o1.length(), o2.length()) && pos1 == pos2; i++) {
                    pos1 = finalKeyword.indexOf(o1.charAt(i));
                    pos2 = finalKeyword.indexOf(o2.charAt(i));
                }

                if (pos1 == pos2 && o1.length() != o2.length()) {
                    return o1.length() - o2.length();
                }

                return pos1  - pos2  ;
            }
        });



 public static void main(String args[]){
       final String order = "!@#$%^ ()*+,-/0123456789=abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ[]_";
customSort(order,  arr);
    }

Answer 1

您的问题有点令人困惑，因为您说您的方法需要验证输入是否有序，但如果不是，则对其进行排序。由于这与排序实际上是相同的，这就是我正在实现的。

您提供的解决方案确实存在一些错误。例如，您在 for 循环中更改 pos1 和 pos2，但仅在 for 循环终止后检查这些值。

在我提供的解决方案中，我构建了一个查找映射来查找字符的顺序，这比在必须进行比较时对字符串执行 indexOf 更有效。此外，每次遇到不在输入顺序或 ALL_CHARS 中的字符时，我们只需将其添加到位置最高的地图中（最后排序）。

public final static String ALL_CHARS = "!@#$%^&()*+,-/0123456789=abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ[]_";

public static void customSort(String order, String[] array) {
    final Map<Character, Integer> orderMap = new HashMap<>();
    String temp = order + ALL_CHARS;
    int len = temp.length();
    for (int i = 0; i < len; i++) {
        orderMap.putIfAbsent(temp.charAt(i), orderMap.size());
    }

    Arrays.sort(array, new Comparator<String>() {

        @Override
        public int compare(String o1, String o2) {
            int min = Math.min(o1.length(), o2.length());
            for (int i = 0; i < min; i++) {
                Integer pos1 = orderMap.get(o1.charAt(i));
                if (pos1 == null) {
                    // character not found, add it to the "end" of our ordering map
                    pos1 = orderMap.size();
                    orderMap.put(o1.charAt(i), pos1);
                }
                Integer pos2 = orderMap.get(o2.charAt(i));
                if (pos2 == null) {
                    // character not found, add it to the "end" of our ordering map
                    pos2 = orderMap.size();
                    orderMap.put(o2.charAt(i), pos2);
                }
                if (pos1 != pos2)
                    // If their different, we now know the ordering
                    return pos1 - pos2;
            }
            // If the start with the same characters, the shortest string should be first
            return o1.length() - o2.length();
        }
    });
}

@Test
public void testCustomSort() {
    String[] test = new String[] { "cat", "cats" };
    customSort("cba", test);
    assertEquals("cat", test[0]);
    customSort("abc", test);
    assertEquals("cat", test[0]);

    test = new String[] { "cat", "dog" };
    customSort("abcd", test);
    assertEquals("cat", test[0]);
    customSort("dcba", test);
    assertEquals("dog", test[0]);

}