# 722. Remove-Comments

difficulty: Medium

&#x20;section pre{ background-color: #eee; border: 1px solid #ddd; padding:10px; border-radius: 5px; }&#x20;

Given a C++ program, remove comments from it. The program `source` is an array where `source[i]` is the `i`-th line of the source code. This represents the result of splitting the original source code string by the newline character `\n`.

In C++, there are two types of comments, line comments, and block comments.

&#x20;The string `//` denotes a line comment, which represents that it and rest of the characters to the right of it in the same line should be ignored.

&#x20;The string `/*` denotes a block comment, which represents that all characters until the next (non-overlapping) occurrence of `*/` should be ignored. (Here, occurrences happen in reading order: line by line from left to right.) To be clear, the string `/*/` does not yet end the block comment, as the ending would be overlapping the beginning.

&#x20;The first effective comment takes precedence over others: if the string `//` occurs in a block comment, it is ignored. Similarly, if the string `/*` occurs in a line or block comment, it is also ignored.

&#x20;If a certain line of code is empty after removing comments, you must not output that line: each string in the answer list will be non-empty.

&#x20;There will be no control characters, single quote, or double quote characters. For example, `source = "string s = "/* Not a comment. */";"` will not be a test case. (Also, nothing else such as defines or macros will interfere with the comments.)

&#x20;It is guaranteed that every open block comment will eventually be closed, so `/*` outside of a line or block comment always starts a new comment.

&#x20;Finally, implicit newline characters can be deleted by block comments. Please see the examples below for details.

After removing the comments from the source code, return the source code in the same format.

**Example 1:**\\

```
Input: 
source = ["/*Test program */", "int main()", "{ ", "  // variable declaration ", "int a, b, c;", "/* This is a test", "   multiline  ", "   comment for ", "   testing */", "a = b + c;", "}"]
The line by line code is visualized as below:
/*Test program */
int main()
{ 
  // variable declaration 
int a, b, c;
/* This is a test
   multiline  
   comment for 
   testing */
a = b + c;
}
Output: ["int main()","{ ","  ","int a, b, c;","a = b + c;","}"]
The line by line code is visualized as below:
int main()
{ 
int a, b, c;
a = b + c;
}
Explanation: 
The string /* denotes a block comment, including line 1 and lines 6-9. The string // denotes line 4 as comments.
```

**Example 2:**\\

```
Input: 
source = ["a/*comment", "line", "more_comment*/b"]
Output: ["ab"]
Explanation: The original source string is "a/*comment\nline\nmore_comment*/b", where we have bolded the newline characters.  After deletion, the implicit newline characters are deleted, leaving the string "ab", which when delimited by newline characters becomes ["ab"].
```

**Note:**

The length of `source` is in the range `[1, 100]`.

The length of `source[i]` is in the range `[0, 80]`.

Every open block comment is eventually closed.

There are no single-quote, double-quote, or control characters in the source code.

## Method One

可以学习一下RegExr <https://leetcode.com/problems/remove-comments/discuss/109195/1-liners>

```java
class Solution {
    public List<String> removeComments(String[] source) {
        ArrayList<String> ans = new ArrayList<>();
        StringBuilder sb = new StringBuilder();
        boolean isBlock = false;
        for(String s : source){
            for(int i = 0; i < s.length(); i++){
                char c = s.charAt(i);
                if(isBlock){
                    // 如果发现在 block 里面，其实我们什么都不用做。
                    // 唯一需要做的就是检测有木有 block 终结的符号。
                    if( i < s.length() - 1 && c == '*' && s.charAt(i+1) == '/'){
                        isBlock = false;
                        i++; // i++ 是为了跳过 不仅仅 * 还有/.
                    }

                }else{
                    // 如果不在 block 里面，只需要注意 // 就可以了
                    // 当然还需要注意 /* 开始一个新的block
                    if( i < s.length() - 1 && c == '/' && s.charAt(i+1) == '*'){
                        isBlock = true;
                        i++; // i++ 是为了跳过 不仅仅 * 还有/.
                        continue;
                    }else if ( i < s.length() - 1 && c == '/' && s.charAt(i+1) == '/'){
                        break;
                    }else{
                        // 注意了，这个题目这里，如果遇到除掉 // 之后，该行是只有空格的情况是不予额外处理的。
                        // 如果要求处理那种情况，这个题就非常难了。但是一般是不要求这么处理的。
                        sb.append(c);
                    }
                }
            }
            if(!isBlock && sb.length() > 0){
                ans.add(sb.toString());
                sb = new StringBuilder();
            }
        }
        return  ans;
    }
}
​
```
