ZigZag Conversion

问题描述：

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   NA P L S I I GY   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".

解决思路：找出规律即可，每次的周期为2*nRows-2,但是在1到nRows-1的行数还需要判断周期为2*（nRows-1-i）的元素，i是处于1到nRows-1的行数。

C++:

class Solution {public:    string convert(string s, int nRows) {        if(s.empty() || nRows <= 1 || s.length() < nRows)            return s;        string result;        int cycle = 2 * nRows - 2;        for(int i = 0; i < nRows; i++){            for(int j = i; j < s.length(); j += cycle){                result.push_back(s[j]);                if( i > 0 && i < nRows-1){                    if (j + 2 *(nRows - 1 - i) < s.length())                        result.push_back(s[j + 2 *(nRows - 1 - i)]);                }            }        }        return result;    }};

python:

class Solution:    # @return a string    def convert(self, s, nRows):        if s == None or len(s) < nRows or nRows < 2:            return s        result = ""        cycle = 2*(nRows-1)        for i in range(0,nRows):            for j in range(i,len(s),cycle):                result += s[j]                if i > 0 and i < nRows-1:                    if (j + 2 *(nRows-1-i)) < len(s):                        result += s[j + 2 *(nRows-1-i)]        return result