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：

把字符串先用二维矩阵存下来，然后先行后列存储到新string即可。注意Z形稀疏矩阵可以压缩存储，使空间复制度还是O（n）。AC代码如下：

 

 1 class Solution {
 2 public:
 3     string convert(string s, int numRows) {
 4         if(s==""|| numRows==1){
 5             return s;
 6         }
 7         int residual,i_col,n=s.size();
 8         int circle=2*numRows-2,count=0;
 9         int col=(n-1)/(numRows-1)+1;
10         char tmp[numRows][col];
11         string res;
12         memset(tmp, 0, sizeof(tmp));
13         for(int i=0;i<n;i++){
14             residual=i%circle;
15             i_col=2*(i/circle)+(i%circle)/numRows;
16             if(residual<numRows)
17                 tmp[residual][i_col]=s[i];
18             else
19                 tmp[numRows-residual%numRows-2][i_col]=s[i];
20         }
21         for(int i=0;i<numRows;i++)
22             for(int j=0;j<col;j++)
23                 if(tmp[i][j]!='\0')
24                     res+=tmp[i][j];
25         return res;
26     }
27 };