【leetcode】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".

思路：

我自己是先用vector把每一行的字母都存下来，再按行读取

string convert(string s, int nRows) {
        string ans = s;
        if(nRows == 1) return ans; //只有一行的特殊对待 没有向上和向下的区别
        vector<vector<char>> v(nRows);
        int z = 0;
        bool add = true;
        for(int i = 0; i < s.length(); i++)
        {
            if(add) //当前行数在增加
            {
                v[z++].push_back(s[i]);
                if(z == nRows)
                {
                    add = false;
                    z = nRows - 2;
                }
            }
            else //当前行数在下降
            {
                v[z--].push_back(s[i]);
                if(z < 0)
                {
                    add = true;
                    z = 1;
                }
            }
        }
        //写入答案
        int p = 0;
        for(int i = 0; i < nRows; i++)
        {
            for(int j = 0; j < v[i].size(); j++)
            {
                ans[p++] = v[i][j];
            }
        }
        return ans;
    }

开始也想了数学的方式，但自己没想清楚。看看别人写的数学方法：

class Solution {
public:
string convert(string s, int nRows) {
    if(s.empty()||nRows<2)
        return s;
    vector<string> zig(nRows);
    bool down = true;
    for(int i = 0; i < s.size(); ++i)
    {
        int row = i%(nRows - 1);
        if(down)
            zig[row].push_back(s[i]);

        else
            zig[nRows - 1 - row].push_back(s[i]);
        if(row==nRows - 2)
            down = !down;
    }
    string res;
    for(auto& temp: zig)
        res += temp;
    return res;
}
};