LeetCode - ZigZag Conversion

Question

Link : https://leetcode.com/problems/zigzag-conversion/

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 text, int nRows);
convert(“PAYPALISHIRING”, 3) should return “PAHNAPLSIIGYIR”.

Code

额，感觉这个比较挫的算法空间复杂度有点高，为O(n)，同时时间复杂度也为O(n)。（C++ : 72ms）

class Solution {
public:
    string convert(string s, int numRows) {
        if(numRows == 1)
            return s;
        vector<vector<char>> rows;
        for(int i = 0; i < numRows; i++){
            vector<char> temp;
            rows.push_back(temp);
        }
        int index = 0, dir = 0; // dir == 0 means down, or up;
        string res;
        for(int i = 0; i < s.size(); i++){
            rows[index].push_back(s[i]);
            if(dir == 0){
                if(++index == (numRows - 1))
                    dir = 1;
            }
            else{
                if(--index == 0)
                    dir = 0;
            }
        }
        for(int i = 0; i < numRows; i++)
            for(int j = 0; j < rows[i].size(); j++)
                res.push_back(rows[i][j]);
        return res;
    }
};

一个高级点的方式，直接计算结果值。（C++ : 20ms）

class Solution {
public:
    string convert(string s, int numRows) {
        string str;
        if(numRows == 1) return s;
        int i, j, k, size = s.size();
        for(i = 0; i < numRows; i++){
            for(k = 0, j = i; j < size;){
                j = (numRows - 1) * k - i;
                if(j >= size) break;
                    str.push_back(s[j]);

                j = (numRows - 1) * k + i;
                if(j >= size) break;
                if(i != 0 && i != (numRows - 1))
                    str.push_back(s[j]);
                k += 2;
            }
        }
        return str;
    }
};