英文:
How to make string comparison inclusive?
问题
我有一个包含单词及其定义的TreeMap。我正在编写一个方法，该方法将返回在范围内的所有单词及其定义；它接受两个参数：string begin 和 string end。
以下是我目前的代码：
我在函数外部初始化了一个TreeMap：
Map<String, List<String>> wordDictionary = new TreeMap<String, List<String>>();
Set<String> wordSet = wordDictionary.keySet();Iterator<String> iterator = wordSet.iterator();while (iterator.hasNext()) {String current = iterator.next();if ((current.compareToIgnoreCase(begin) >= 0) &&(current.compareToIgnoreCase(end) <= 0)) {List<String> defList = wordDictionary.get(current);System.out.println(current);Iterator<String> itr2 = defList.iterator();while (itr2.hasNext()) {System.out.println(" " + itr2.next());}}}
示例用法： "List a b"
该方法应返回从 a 到 b 的所有单词。
然而，我的方法返回的是在 a 和 b 之间的所有单词；它不包括以 a 和 b 开头的单词。
我尝试调整 if 语句为
if ((current.compareToIgnoreCase(begin) > 0) &&(current.compareToIgnoreCase(end) < 0))
但那也不起作用。它会显示以 begin 开头但不以 end 开头的单词。希望这能理解。
英文:
I have a TreeMap that contains words and its definition(s). I'm writing a method that will return all the word and its definitions with a range; it takes in two parameters: string begin and string end.
Here's what I have so far
I initialized a TreeMap outside of the function:
Map<String, List<String>> wordDictionary = new TreeMap<String, List<String>>();
Set<String> wordSet = wordDictionary.keySet();Iterator<String> iterator = wordSet.iterator();while (iterator.hasNext()) {String current = iterator.next();if ((current.compareToIgnoreCase(begin) >= 0) &&(current.compareToIgnoreCase(end) <= 0)) {List<String> defList = wordDictionary.get(current);System.out.println(current);Iterator<String> itr2 = defList.iterator();while (itr2.hasNext()) {System.out.println(" " + itr2.next());}}}
Example use: "List a b"
The method should return all words from a to b <br>
However my method returns all words between a and b; it excludes words starting with a and b.
I tried tweaking the if case to
if ((current.compareToIgnoreCase(begin) > 0) &&(current.compareToIgnoreCase(end) < 0))
but that doesn't work either. It displays words starting with begin but not end. Hope that makes sense.
答案1
得分: 3
如果你有一个 TreeMap，使用它的 subMap() 方法！
Map<String, List<String>> range = wordDictionary.subMap(begin, true, end, true);
然后遍历整个映射。
第 2个 和第 4个 参数控制范围的结束是否包含在内 - 看起来你想要两端都包含，但如果不是这样，将第 4个 参数改为 false。
——
顺便说一下，由于你不是打印单词，只是定义，你的代码可以简化为：
wordDictionary.subMap(begin, true, end, true).values().stream().flatMap(List::stream).forEach(System.out::println);
如果你想在缩进的定义之前也打印单词：
wordDictionary.subMap(begin, true, end, true).entrySet().stream().peek(e -> System.out.println(e.getKey())).map(Map.Entry::getValue).flatMap(List::stream).forEach(d -> System.out.println(" " + d));
英文:
If you have a TreeMap, use its subMap() method!
Map<String, List<String>> range = wordDictionary.subMap(begin, true, end, true);
And iterate over the whole map.
The 2<sup>nd</sup> and 4<sup>th</sup> parameters control if the range ends are inclusive or not - it seems you want inclusive both ends, but if not change the 4<sup>th</sup> parameter to false.
——
By the way, since you are not printing the word, just the definitions, your code can be reduced to:
wordDictionary.subMap(begin, true, end, true).values().stream().flatMap(List::stream).forEach(System.out::println);
If you wanted to also print the word before indented definitions:
wordDictionary.subMap(begin, true, end, true).entrySet().stream().peek(e -> System.out.println(e.getKey())).map(Map.Entry::getValue).flatMap(List::stream).forEach(d -> System.out. println(" " + d);
答案2
得分: 2
你可以通过类似以下方式使TreeMap的subMap在两端都是闭区间：
TreeMap<String, String> map = new TreeMap<>();map.put("abc", "a");map.put("efg", "a");map.put("hij", "v");map.put("rst", "o");String start = "a";String end = "e";int idx = end.length() - 1;// 修改结束字符串的最后一个字符为包含在内。StringBuilder sb = new StringBuilder(end);char c = sb.charAt(sb.length() - 1);sb.setCharAt(sb.length() - 1, (char) (c + 1));end = sb.toString();Map<String, String> words =map.subMap(start, true, end, false);System.out.println(words);
输出
{abc=a, efg=a}
这会递增结束字符串的最后一个字符，以包含小于那个字符的任何内容。
英文:
You could make the subMap of TreeMap inclusive on both ends by doing something like the following:
TreeMap<String,String> map = new TreeMap<>();map.put("abc","a");map.put("efg","a");map.put("hij","v");map.put("rst","o");String start = "a";String end = "e";int idx = end.length() - 1;// modify last character of end to be inclusive.StringBuilder sb = new StringBuilder(end);char c = sb.charAt(sb.length()-1);sb.setCharAt(sb.length()-1, (char)(c+1));end = sb.toString();Map<String, String> words =map.subMap(start, true, end,false);System.out.println(words);
Prints
{abc=a, efg=a}
It increments the last character of the ending string so as to include anything less than that.
答案3
得分: 2
获取所有以'b'开头的键，使用 subMap(K fromKey, boolean fromInclusive, K toKey, boolean toInclusive) 方法：
TreeMap<String, String> wordDictionary = new TreeMap<>(String.CASE_INSENSITIVE_ORDER);wordDictionary.put("A", "1");wordDictionary.put("Abc", "2");wordDictionary.put("Axe", "3");wordDictionary.put("B", "4");wordDictionary.put("Bee", "5");wordDictionary.put("C", "6");wordDictionary.put("Car", "7");NavigableMap<String, String> bdict = wordDictionary.subMap("b", true, "c", false);System.out.println(bdict);System.out.println(bdict.keySet());// AlternativeNavigableSet<String> bwords = wordDictionary.navigableKeySet().subSet("b", true, "c", false);System.out.println(bwords);
输出
{B=4, Bee=5}[B, Bee][B, Bee]
英文:
To get all keys that start with e.g. 'b', use subMap(K fromKey, boolean fromInclusive, K toKey, boolean toInclusive):
TreeMap<String, String> wordDictionary = new TreeMap<>(String.CASE_INSENSITIVE_ORDER);wordDictionary.put("A", "1");wordDictionary.put("Abc", "2");wordDictionary.put("Axe", "3");wordDictionary.put("B", "4");wordDictionary.put("Bee", "5");wordDictionary.put("C", "6");wordDictionary.put("Car", "7");NavigableMap<String, String> bdict = wordDictionary.subMap("b", true, "c", false);System.out.println(bdict);System.out.println(bdict.keySet());// AlternativeNavigableSet<String> bwords = wordDictionary.navigableKeySet().subSet("b", true, "c", false);System.out.println(bwords);
Output
{B=4, Bee=5}[B, Bee][B, Bee]
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