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".

 

思路：

1、numRows可以计算出每多少个字母一个循环（例如，上面例子中为4个字母一个循环）

2、竖直的字符个数为numRows，斜杠部分的字符个数实际上是numRows - 2，每(numRows * 2 - 2)个数字一个循环

3、每次偏移量增加(numRows * 2 - 2)，依次输出即可。判断该位置是否有字符的唯一标准为是否越界，so easy

 

源代码如下：

#include 
     
      #include 
      
       #include 
       
        using namespace std;class Solution {public:    string convert(string s, int numRows) {        if (numRows == 1) return s;        string strResult;        int iLen = s.size();        int iCellNum = numRows * 2 - 2;        for (int i = 0; i < numRows; ++i)        {            if (i == 0 || i == numRows - 1)            {                int j = i;                while (j < iLen)                {                    strResult += s[j];                    j += iCellNum;                }            }            else            {                int j = i;                int k = iCellNum - i;                while (j < iLen)                {                    strResult += s[j];                    j += iCellNum;                    if (k < iLen)                    {                        strResult += s[k];                        k += iCellNum;                    }                }            }        }        return strResult;    }};int main(){    Solution a;    string strResult = a.convert("PAYPALISHIRING", 3);    printf("%s\n", strResult.c_str());    system("pause");    return 0;}