java8 summarizes the of recommendation by several ways to count the number of letters in a string
- 2020-11-25 07:16:33
- OfStack
1. Count the number of letters in the string (and keep them in alphabetical order)
aabbbbbbbba oh, bcab cdabc deaaa
There are 5 ways I know how to do it so far. Oh, if you have better ones, please tell me
// way 1
public static void letterCount1(String s) {
s=s.replaceAll(" +", "");
//1, Convert to an array of characters
char c[]=s.toCharArray();
Map<Character, Integer> tree=new TreeMap<Character, Integer>();
for (int i = 0; i < c.length; i++) {
// The first 1 Time: a,1
// The first 2 Time: a,2
//2, Gets the value corresponding to the key
Integer value=tree.get(c[i]);
// The compiler: Integer value = (Integer)tree.get(Character.valueOf(c[i]));
//3, Storage judgment
tree.put(c[i], value==null? 1:value+1);
}
// If the result format is required: a(5)b(4)c(3)d(2)e(1)
StringBuilder sbu=new StringBuilder();
for(Character key:tree.keySet()){
Integer count=tree.get(key);
sbu.append(key).append("(").append(count).append(")");
}
// will sbu Convert to string
System.out.println(sbu.toString());
}
// way 2 Use the stream
// This is testing special characters , Such as \ \n when , He's not in the right order , This is a Map Caused by the
// Solution use TreeMap
public static void letterCount2(String s) {
s=s.replaceAll(" +", "");
TreeMap<String, Long> result = Arrays.stream(s.split(""))
.sorted()
// .collect(Collectors.groupingBy(Function.identity(),Collectors.counting()));
.collect(Collectors.groupingBy(Function.identity(),TreeMap::new,Collectors.counting()));
System.out.println(result);
}
// way 3 use Collections.frequency
// It's just a string that becomes a set for each string , Compare each string loop to a collection
public static void letterCount3(String s) {
s=s.replaceAll(" +", "");
List<String> list=Arrays.asList(s.split(""));
Map<String,Integer> map=new TreeMap<String, Integer>();
for (String str : list) {
map.put(str, Collections.frequency(list, str));
}
System.out.println(map);
}
// way 4
public static void letterCount4(String s) {
s=s.replaceAll(" +", "");
String[] strs = s.split("");
Map<String,Integer> map=new TreeMap<String, Integer>();
for (String str : strs) {
map.put(str, stringCount(s, str));
}
System.out.println(map);
}
// way 5
public static void letterCount5(String s) {
s=s.replaceAll(" +", "");
String[] strs = s.split("");
Map<String,Integer> map=new TreeMap<String, Integer>();
for (String str : strs) {
map.put(str, stringCount2(s, str));
}
System.out.println(map);
}
// Use opportunely split
public static int stringCount(String maxstr, String substr) {
// Pay attention to
// 1. Such as qqqq, Could not find , Returns the string directly
// 2. Such as qqqjava, There are no other characters at the end , There's no splitting , So you can add 1 A blank space
// 3.java11 No characters at the beginning, no relationship, automatic empty fill
// 4. For special characters , Be careful with escape characters
int count = (maxstr + " ").split(substr).length - 1;
// System.out.println("\"" + minstr + "\"" + " Occurrence times of string: " + count);
return count;
}
// If case insensitive, then compile(minstr,CASE_INSENSITIVE)
public static int stringCount2(String maxstr, String substr) {
int count = 0;
Matcher m = Pattern.compile(substr).matcher(maxstr);
while (m.find()) {
count++;
}
return count;
}
2. Count the number of words in the string (English only)
This is actually the same thing as the 1 up here, but let me just write 1 neat way to do it
public static void wordStringCount(String s) {
// So here we start with a string , The split becomes a stream of strings
Map<String, Long> result = Arrays.stream(s.split("\\s+"))
.map(word -> word.replaceAll("[^a-zA-Z]", ""))
.collect(Collectors.groupingBy(Function.identity(),Collectors.counting()));
System.out.println(result);
}
3. Count the number of words in the text (English only)
// statistical 1 The number of words in the text
public static void wordFileCount(String path) throws IOException{
// Here, 1 Start string stream
// First division
// In becoming a character stream
// In the screen
Map<String, Long> result = Files.lines(Paths.get(path),Charset.defaultCharset())
.parallel()
// String flow -- segmentation -- String flow
.flatMap(str->Arrays.stream(str.split(" +")))
.map(word -> word.replaceAll("[^a-zA-Z]", ""))
// Remove empty
.filter(word->word.length()>0)
.collect(Collectors.groupingBy(Function.identity(),Collectors.counting()));
System.out.println(result);
}
4. Other irrelevant items
We know that mutable argument list, you can pass no arguments
for
public void testName() {
System.out.println("a");
}
public void testName(String ... s) {
// Don't pass parameters ,s Will be initialized by default 1 An object
System.out.println("b");
}
Now call testName() to print what ? , which prints a and automatically matches methods whose parameters are really empty