java - UVa #494 - 正则表达式 [^a-zA-z]+ 使用 Java 分割单词

Question

我正在玩UVa #494我设法用下面的代码解决了这个问题:

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

class Main {    
    public static void main(String[] args) throws IOException{
        BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
        String line;
        while((line = in.readLine()) != null){
            String words[] = line.split("[^a-zA-z]+");
            int cnt = words.length;
            // for some reason it is counting two words for 234234ddfdfd and words[0] is empty
            if(cnt != 0 && words[0].isEmpty()) cnt--; // ugly fix, if has words and the first is empty, reduce one word
            System.out.println(cnt);
        }
        System.exit(0);
    }
}

我构建了正则表达式 "[^a-zA-z]+" 来分割单词，例如字符串 abc..abc 或 abc432abc 应拆分为 ["abc", "abc"]。但是，当我尝试字符串 432abc 时，结果是 ["", "abc"] - 来自 words[] 的第一个元素> 只是一个空字符串，但我期望只有 ["abc"]。我不明白为什么这个正则表达式给我第一个元素作为本例的 "" 。

Answer 1

查看拆分引用页:split reference

Each element of separator defines a separate delimiter character. If two delimiters are adjacent, or a delimiter is found at the beginning or end of this instance, the corresponding array element contains Empty. The following table provides examples.

由于您有多个连续的分隔符，因此您将得到空数组元素