[Leetcode] 6. Zigzag Conversion

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The string "PAYPALISHIRING" is written in a zigzag pattern on a given number of rows like this: (you may want to display this pattern in a fixed font for better legibility)

P   A   H   N
A P L S I I G
Y   I   R

And then read line by line: "PAHNAPLSIIGYIR"

Write the code that will take a string and make this conversion given a number of rows:

string convert(string s, int numRows);

Example 1:

Input: s = "PAYPALISHIRING", numRows = 3
Output: "PAHNAPLSIIGYIR"

Example 2:

Input: s = "PAYPALISHIRING", numRows = 4
Output: "PINALSIGYAHRPI"
Explanation:
P     I    N
A   L S  I G
Y A   H R
P     I

Example 3:

Input: s = "A", numRows = 1
Output: "A"

Constraints:

• 1 <= s.length <= 1000
• s consists of English letters (lower-case and upper-case), ',' and '.'.
• 1 <= numRows <= 1000

Solution

line별로 문자열을 넣어두는 list를 만드는데, 만드는 방법은 맨 밑과 윗줄에 닿았을때만 방향을 바꿔주면 되는데 방향을 바꾼다는 의미가 currentLine을 더하거나 빼거나를 정해준다는 뜻. 

if 맨 윗줄이거나 맨 아래에 닿으면 -> 방향부호를 정함

아래로 내려가면 -> line ++

위로 올라가면 -> line --

class Solution {
public:
    string convert(string s, int numRows) {
        if(numRows<=1) return s;
        string ret="";
        vector<string> list(numRows, "");
        int currentLine = 0;
        bool reachedEdge = true;

        for (int i = 0; i < s.length(); ++i) {
            if (currentLine == 0 || currentLine == numRows - 1) {
                reachedEdge = !reachedEdge;
            }
            list.at(currentLine) += s.at(i);
            if (!reachedEdge) {
                currentLine++;
            } else {
                currentLine--;
            }
        }
        for(int i=0;i<numRows;i++)
            ret+=list.at(i);
        return ret;
    }
};