Question

您能否看一下我的代码，并告诉我为什么它不能按预期工作-

   String word1 = "Simple sentence that is first";
    String word2 = "Another sentence that is second.";
    appear(word1, word2);
}
static void appear(String word1, String word2){
    String[] split = word1.split(" ");
    String[] split2 = word2.split(" ");
    for (int i = 0; i <= split.length - 1; i++){
        for (int j = 0; j <= split2.length - 1; j++){
            if (!split[i].equals(split2[j])){
                System.out.println(split[i]);

            }
        }
    }
}

输出应该是：Simple和first，因为它们是第二个句子中没有出现的仅有的两个词。我的输出是第一句话中的所有单词重复了几次。（！.equals）应该工作吗？

Answer 1

通过这种方式，您正在检查第一个句子的每个单词是否等于第二个句子的每个单词。举例来说，这意味着对于句子一词，它将检查它是否与另一个相等，如果不相等，将进行porint，然后将其检查是否与句子相等，依此类推。如果要打印，则应跟踪第二句中是否出现第一句话的单词

String word1 = "Simple sentence that is first";
String word2 = "Another sentence that is second.";
appear(word1, word2);
}
static void appear(String word1, String word2){
    String[] split = word1.split(" ");
    String[] split2 = word2.split(" ");
    boolean appeared;
    for (int i = 0; i <= split.length - 1; i++){
        appeared = false;
        for (int j = 0; j <= split2.length - 1; j++){
            if (split[i].equals(split2[j])){
                appeared = true;
            }
        }
    if(appeared == false)
        System.out.println(split[i]);
    }
}

Answer 2

字符串具有contains（）方法，您可以依靠它搜索第一句中的每个单词是否出现在第二句中。

    String[] split = word1.split(" ");
    for (int i = 0; i <= split.length - 1; i++){
    if(!Arrays.asList(word2.split(" ")).contains(split[i])){
      System.out.println(split[i]);
    }
    }

使用Java8流，可以将上述内容修改为以下一行。

Arrays.asList(word1.split(" ")).stream().filter(x->!Arrays.asList(word2.split(" ")).contains(x)).forEach(System.out::println);

Answer 3

解决问题的关键是一个称为Set的数据结构。 GeeksForGeeks的介绍很好。

使用集合，我们可以将问题建模为两组字符串之间的差异。我试图与Geeks for Geeks文章一致地命名变量，以便您可以继续学习。

Set<String> a = new HashSet<>(); 
Set<String> b = new HashSet<>();
a.addAll(word1.split(" "));
// a is now {"Simple" "sentence" "that" "is", "first"}
b.addAll(word2.split(" "));
// b is now {"Another", "sentence", "that", "is", "second";

// We can now calculate the difference
a.removeAll(b);
// a is now {"Simple", "first"}

// Because sets are iterable we can finish via a for-each loop
for(String s : a) {
    System.out.println(s);
}

输出第二句中未出现的所有单词

3 个答案: