删除数组中重复的单词句子

Question

给定一个决定单词输出、每个单词的长度和单词重复次数的问题，我有以下代码能够确定下面每个单词的单词和长度：

String sentence;
String charSentence;
String[] wordOutput;
private void analyzeWords(String s) {
String[] words = sentence.split(" ");
wordOutput = new String[words.length];
int[] repeats = new int[words.length];
// Increment a single repeat
for (int i = 0; i < words.length; i++) {
repeats[i] = 1;
// Increment when repeated.
for (int j = i + 1; j < words.length - 1; j++) {
if (words[i].equalsIgnoreCase(words[j])) {
repeats[i]++;
}
}
wordOutput[i] = words[i] + "t" + words[i].length() + "t" + repeats[i];
}

当我运行程序时，我得到以下输出：

Equal   5   2
Equal   5   1 <- This is a duplicate word and should not be here when it repeats.

有人知道我的问题在哪里吗？是不是与我的重复数组有关？

Answer 1

第一个问题是，在内部for循环中，您正在从i+1循环到length-1。你需要循环到length。其次，您需要确定该单词是否在String中出现，如果出现，请使用continue语句。你可以做：

outer:
for (int i = 0; i < words.length; i++) {
repeats[i] = 1;
for(int index = i-1; index >= 0; index--) {
if(words[i].equals(words[index])) {
continue outer;
}
}
...
}

然而，这样做的问题是，当您指定一个长度与字数相同的Array时，列表末尾会有null值。要解决这个问题，你可以做：

wordOutput = Arrays.stream(wordOutput).filter(e-> e!= null).toArray(String[]::new);

它将过滤掉null值

输出：

(输入String:"This is a String is a with a lot lot of this repeats repeats"(

This    4   2
is      2   2
a       1   3
String  6   1
with    4   1
lot     3   2
of      2   1
this    4   1
repeats 7   2

Answer 2

不是在所有索引中递增计数，而是只在单词的最后一次出现时存储计数，在其他情况下，计数值为0。最后遍历计数数组，如果它大于零，则打印值及其计数

private void analyzeWords(String s) {
String[] words = sentence.split(" ");
wordOutput = new String[words.length];
int[] repeats = new int[words.length];
for (int i = 0; i < words.length; i++) {
int count =1;
int index = i;
for (int j = i + 1; j < words.length - 1; j++) {
if (words[i].equalsIgnoreCase(words[j])) {
count++;
index = j;
}
}
if(repeats[index]==0){
repeats[index]=count; // update repeat array only for last occurence of word
wordOutput[i] = words[i] + "t" + words[i].length() + "t" + repeats[index];
}
}

Answer 3

首先，正如GBlodgett提到的，您应该检查所有剩下的单词是否重复，您当前的解决方案跳过最后一个单词。将第二个循环终止条件更新为j < words.length。

其次，如果只想打印一次重复项，则需要在解决方案中设置一个条件。其中一个例子：

boolean[] duplicates = new boolean[words.length];
// Increment a single repeat
for (int i = 0; i < words.length; i++) {
repeats[i] = 1;
// Check for duplicates,
// If the word was not marked as duplicate
if (!duplicates[i]) {
// Increment when repeated.
for (int j = i + 1; j < words.length; j++) {
if (words[i].equalsIgnoreCase(words[j])) {
repeats[i]++;
duplicates[j] = true;
}
}
wordOutput[i] = words[i] + "t" + words[i].length() + "t" + repeats[i];
}
}

---

有一个Java 8+解决方案，例如：

Map<String, Long> m = Arrays.stream(s.split(" ")).collect(Collectors.groupingBy(Function.identity(), Collectors.counting()));

地图将出现成对的单词和it。