Remove Comments

https://leetcode.com/contest/leetcode-weekly-contest-57/problems/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:

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.

Thoughts

原先想复杂了, 在block里无论是不是"//"都忽略, 只要判断是不是在block里, 而且只有不在block时才重新new 出一个sb.

Code

class Solution {
    public List<String> removeComments(String[] source) {
        List<String> res = new ArrayList<>();
        boolean inBlock = false;
        StringBuilder sb = new StringBuilder();
        for (String line : source) {
            for (int i = 0; i < line.length(); i++) {
                char c = line.charAt(i);
                if (!inBlock) {
                    if (i != line.length() - 1 && c == '/' && line.charAt(i + 1) == '/') {
                        break;
                    } else if (i != line.length() - 1 && c == '/' && line.charAt(i + 1) == '*') {
                        inBlock = true;
                        i++;
                    } else {
                        sb.append(c);
                    }
                } else {
                    if (i != line.length() - 1 && c == '*' && line.charAt(i + 1) == '/') {
                        inBlock = false;
                        i++;
                    }
                }
            }
            if (!inBlock && sb.length() > 0) {
                res.add(sb.toString());
                sb = new StringBuilder();
            }
        }
        return res;
    }
}

Analysis

时空复杂度O(N)