https://leetcode.com/problems/remove-comments/description/
The length of
The length of
Every open block comment is eventually closed.
There are no single-quote, double-quote, or control characters in the source code.
https://leetcode.com/articles/remove-comments/
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.
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.
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.
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.
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.
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.)
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.
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:
source
is in the range [1, 100]
.source[i]
is in the range [0, 80]
.https://leetcode.com/articles/remove-comments/
We need to parse the
source
line by line. Our state is that we either are in a block comment or not.- If we start a block comment and we aren't in a block, then we will skip over the next two characters and change our state to be in a block.
- If we end a block comment and we are in a block, then we will skip over the next two characters and change our state to be not in a block.
- If we start a line comment and we aren't in a block, then we will ignore the rest of the line.
- If we aren't in a block comment (and it wasn't the start of a comment), we will record the character we are at.
- At the end of each line, if we aren't in a block, we will record the line.
public List<String> removeComments(String[] source) { boolean inBlock = false; StringBuilder newline = new StringBuilder(); List<String> ans = new ArrayList(); for (String line: source) { int i = 0; char[] chars = line.toCharArray(); if (!inBlock) newline = new StringBuilder(); while (i < line.length()) { if (!inBlock && i+1 < line.length() && chars[i] == '/' && chars[i+1] == '*') { inBlock = true; i++; } else if (inBlock && i+1 < line.length() && chars[i] == '*' && chars[i+1] == '/') { inBlock = false; i++; } else if (!inBlock && i+1 < line.length() && chars[i] == '/' && chars[i+1] == '/') { break; } else if (!inBlock) { newline.append(chars[i]); } i++; } if (!inBlock && newline.length() > 0) { ans.add(new String(newline)); } } return ans; }
https://leetcode.com/problems/remove-comments/discuss/109197/One-pass-solution-in-Java
https://leetcode.com/problems/remove-comments/discuss/109195/1-liners
We only need to check for two things:
- If we see '//' we stop reading the current line, and add whatever characters we have seen to the result.
- If we see '/*' then we start the multiline comment mode and we keep on ignoring characters until we see '*/'.
- If the current character is neither of the above two and the multiline comment mode is off, then we add that character to the current line.
Once we parse one line (source[i]), then if the mode is off, we add the currently generated line (StringBuilder) to the result and repeat for source[i + 1].
We need to be careful not to insert empty lines in the result.
class Solution {
public List<String> removeComments(String[] source) {
List<String> res = new ArrayList<>();
StringBuilder sb = new StringBuilder();
boolean mode = false;
for (String s : source) {
for (int i = 0; i < s.length(); i++) {
if (mode) {
if (s.charAt(i) == '*' && i < s.length() - 1 && s.charAt(i + 1) == '/') {
mode = false;
i++; //skip '/' on next iteration of i
}
}
else {
if (s.charAt(i) == '/' && i < s.length() - 1 && s.charAt(i + 1) == '/') {
break; //ignore remaining characters on line s
}
else if (s.charAt(i) == '/' && i < s.length() - 1 && s.charAt(i + 1) == '*') {
mode = true;
i++; //skip '*' on next iteration of i
}
else sb.append(s.charAt(i)); //not a comment
}
}
if (!mode && sb.length() > 0) {
res.add(sb.toString());
sb = new StringBuilder(); //reset for next line of source code
}
}
return res;
}
}
X. Regulare Expression
public List<String> removeComments(String[] source) { String[] s = String.join("\n",source).replaceAll("//.*|/\\*(.|\n)*?\\*/", "").split("\n"); List<String> ans = new ArrayList<>(); for(String str : s) if(!str.equals("")) ans.add(str); return ans; }