【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;

}

};