Question

我编写了功能代码，用于读取和写入数据值到arraylist。虽然它会返回所有内容，但如果某个值有多个单词，我该如何排除第一个？

  // 3 points
static ArrayList<String> Q2(String filename) {

    // You are given a file (filename) containing a different random phrase on each line. Return an
    // ArrayList containing each phrase, but without the first word of each phrase.
    //
    // Example: If the files contains the 2 phrases "roofed crossover" and "beneficiary charles frederick worth" the
    // ArrayList should contain "crossover" and "charles frederick worth"
    ArrayList<String> al = new ArrayList<String>();

    try {
        for(String s : Files.readAllLines(Paths.get(filename))){


            al.add(s.substring(9));
        }
    } catch (IOException e) {
        // TODO Auto-generated catch block
        e.printStackTrace();
    }

    return al;
}

继承人说的是：

Incorrect on input: data/phrases0.txt
Expected output : [algae, blood platelet, charles frederick worth, convert, crossover, eye movement, ferocity, itch, lake albert, loewi, mountainside, peach, sontag, specialty, supposition, surprised endometriosis, testimonial, trial golden fleece, waterproofing, wrongdoer]
Your output     : [ferocity, peach, ed algae, wi, ossover, ry charles frederick worth, ised endometriosis, wrongdoer,  lake albert, ng waterproofing, d eye movement, mountainside, g testimonial, c itch, tal sontag, ive blood platelet, golden fleece, ic specialty, convert, s supposition]

我已经让它在没有第一个字符串的情况下返回一些值，但有些字大于子字符串可以达到的值。

Answer 1

在迭代拆分字符串

时，忽略第一项可能最容易

    data = "";
    for(String s : Files.readAllLines(Paths.get(filename))){    
         line = s.split(",");
         for (int i = 1; i < line.length; i++) {
             String data = line[i] + System.getProperty("line.separator");
             list.add(data);
         }
    }

根据评论，你的输出中也需要逗号。

Answer 2

你需要的只是删除空格前的第一个单词，然后使用这个代码：从空格char的位置获取输入字符串的子字符串。

for(String s : Files.readAllLines(Paths.get(filename))){
    al.add(s.substring(s.indexOf(" ")+1)));
}

找到第一个空格字符的索引？

2 个答案: