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我今天有一个面试练习：

返回一个排序数组（不区分大小写）。排序后的数组将被排序

 * alphabetically by the first 3 characters, then numerically by 
 * the following number and then alphabetically by the remaining characters with
 * spaces above characters.


import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.Comparator;

import org.junit.Test;

public class MySort {


    public String[] testSortArray(String[] input){
        // TODO: Sort the array

    }

    @Test
    public void testSort() {
        String[] input = new String[8];
        input[0] = "AIR1";
        input[1] = "AIR20b";
        input[2] = "BIR5A";
        input[3] = "AIR20AB";
        input[4] = "AIR10ab";
        input[5] = "AIR2 A";
        input[6] = "AIR111";
        input[7] = "AIR1Z";

        MySort sortTest = new MySort();
        String[] output = sortTest.testSortArray(input);

        String[] expected = new String[8];
        expected[0] = "AIR1";
        expected[1] = "AIR1Z";
        expected[2] = "AIR2 A";
        expected[3] = "AIR10ab";
        expected[4] = "AIR20AB";
        expected[5] = "AIR20b";
        expected[6] = "AIR111";
        expected[7] = "BIR5A";
        assertEquals(Arrays.asList(output), Arrays.asList(expected));

        for (String item : output) {
            System.out.println(item);
        }
    }
}

我已将 testSortArray(String[] 输入实现为：

public String[] testSortArray(String[] input){

        Collections.sort(Arrays.asList(input), new Comparator<String>() {
            public int compare(String o1, String o2) {
                return extractNumber(o1) - extractNumber(o2);
            }

            int extractNumber(String s) {
                String num = s.replaceAll("\\D", "");
                // return 0 if no digits found
                return num.isEmpty() ? 0 : Integer.parseInt(num);
            }
        });
        return input;
    }

你能告诉我我的代码有什么问题吗？谢谢

4

1 回答 1

3

您的比较逻辑显然与您的规格不符。它完全忽略了前三个字符，并忽略了前三个字符之后的数字之后的所有内容。显然，您需要考虑这些事情，否则您永远无法匹配您的规范：

    public int compare(String o1, String o2) {
            String s1 = o1.substring(0, 3);
            String s2 = o2.substring(0, 3);
            if(!s1.equals(s2)) {
                return s1.compareTo(s2);
            }
            String[] fields1 = o1.substring(3).split("[^0-9]", 2);
            String[] fields2 = o2.substring(3).split("[^0-9]", 2);
            int i1 = Integer.parseInt(fields1[0]);
            int i2 = Integer.parseInt(fields2[0]);
            if(i1 != i2) {
                return i1 - i2;
            }
            String r1 = "";
            if(fields1.length > 1) {
                r1 = fields1[1];
            }
            String r2 = "";
            if(fields2.length > 1) {
                r2 = fields2[1];
            }
            return r1.compareTo(r2);
        }

这将符合您的规格。我测试了一个获得以下输出：

AIR1
AIR1Z
AIR2 A
AIR10ab
AIR20AB
AIR20b
AIR111
BIR5A
于 2013-08-13T02:11:45.053 回答