Java嵌套For-loop突然崩溃

Question

我是一名程序员爱好者，并且一直在学习Java大约一个月。所以，我决定接受通过r/dailyprogramming提供的问题。链接下面是对于那些有兴趣：

http://www.reddit.com/r/dailyprogrammer/comments/2nynip/2014121_challenge_191_easy_word_counting/

到目前为止，我已经分裂的话到名为分裂的字符串数组。这些单词全部被降低了下来，句点，逗号和其他一些常见的标点符号，导致数组填充了小写字母词。目前我试图通过取数组中的第一个非空的单词来计算每个单词的出现次数，然后检查每个单元并计算每个单词的出现次数。我使用嵌套for循环和if语句来完成此操作。但是，程序会突然停止而不会返回任何错误。我希望有人能向我解释为什么我的代码突然停止。

一切工作正常，直到这部分的代码。

for (int i = 0; i < splitted.length; i++) { 

    if (splitted[i] != null) { 

     word = splitted[i]; 
     System.out.println("Word is: " + word); 

      for (int j = i; j < splitted.length; j++) { 

       if (splitted[j].contains(word)) { 

        splitted[j] = null; 
        count++; 
       } 
      } 

     System.out.println(word + ": " + count); 
     count = 0; 
    }   
} 

这是在不同点输出的修改代码。我检查了数组的长度，它不是没有限制的。

for (int i = 0; i < splitted.length; i++) { 

    if (splitted[i] != null) { 

     word = splitted[i]; 
     System.out.println("Word is: " + word); 

      for (int j = i; j < splitted.length; j++) { 

       System.out.printf("%d %s %B%n", j, splitted[j], splitted[j].contains(word)); 
       if (splitted[j].contains(word)) { 

        splitted[j] = null; 
        count++; 
       } 

       System.out.println(j + " is less than " + splitted.length); 
      } 

     System.out.println(word + ": " + count); 
     count = 0; 
    } 
     System.out.println(splitted[i] + " " + i);   
} 

被修改为更清晰：问题是检查阵列中的一个空元素之后，突然停止尽管j小于splitted.length少该程序。

输出：

Today was great hello stupid Today. Today was bad. Today was amazing. He is great. He was bad. Now he is great! 
Word is: today 
0 today TRUE 
0 is less than 22 
1 was FALSE 
1 is less than 22 
2 great FALSE 
2 is less than 22 
3 hello FALSE 
3 is less than 22 
4 stupid FALSE 
4 is less than 22 
5 today TRUE 
5 is less than 22 
6 today TRUE 
6 is less than 22 
7 was FALSE 
7 is less than 22 
8 bad FALSE 
8 is less than 22 
9 today TRUE 
9 is less than 22 
10 was FALSE 
10 is less than 22 
11 amazing FALSE 
11 is less than 22 
12 he FALSE 
12 is less than 22 
13 is FALSE 
13 is less than 22 
14 great FALSE 
14 is less than 22 
15 he FALSE 
15 is less than 22 
16 was FALSE 
16 is less than 22 
17 bad FALSE 
17 is less than 22 
18 now FALSE 
18 is less than 22 
19 he FALSE 
19 is less than 22 
20 is FALSE 
20 is less than 22 
21 great FALSE 
21 is less than 22 
today: 4 
null 0 
Word is: was 
1 was TRUE 
1 is less than 22 
2 great FALSE 
2 is less than 22 
3 hello FALSE 
3 is less than 22 
4 stupid FALSE 
4 is less than 22 

感谢，

Answer 1

的问题是，你不要在你的第二个for -loop检查null：

for (int j = i; j < splitted.length; j++) { 
    if (splitted[j].contains(word)) {//what if splitted[j] is null? 
     splitted[j] = null; 
     count++; 
    } 
} 

如果null遇到，例如因为之前的迭代已经将项目设置为null，所以得到NullPointerException。

所以，你应该使用：

for (int j = i; j < splitted.length; j++) { 
    if (splitted[j] != null && splitted[j].contains(word)) {//what if splitted[j] is null? 
     splitted[j] = null; 
     count++; 
    } 
} 

但是也有与此代码是不完全正确的一些其他方面：

• 你应该使用equals，而不是包含因为"foobarqux".contains("bar")是true。
• 您可以通过在计数之前首先对值进行排序，或者使用HashMap<String,Integer>来计算实例来更有效地执行此操作（在O（n log n））时间。

检查此jdoodle。