题目：
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".
解决：
1. 要知道zigzag形状。
2. 周期计算位置，注意特殊位置。
/**
* 把一字符串排成Zigzag形状，然后按行重新组起来。通过计算周期性位置。
* Zigzag形状：
* # #
* # # # #
* # # # #
* # #
* @author Administrator
*
*/
public class ZigzagConvert {
public String convert(String s, int nRows) {
int len = s.length();
if (len <= 1 || nRows == 1)
return s;
int zouqi = nRows > 2 ? (nRows + nRows - 2) : nRows;//计算周期，一竖一上折
StringBuffer sb = new StringBuffer();
for (int i = 0; i < nRows; i++) {//遍历行
int idx = i;
int idxy = nRows + nRows - 2 - i;
while (true) {
if (idx >= len)
break;
sb.append(s.charAt(idx));//一竖
idx += zouqi;
if (i > 0 && idxy > nRows - 1) {//一上折，可能不存在
if (idxy >= len)
break;
sb.append(s.charAt(idxy));
idxy += zouqi;
}
}
}
return sb.toString();
}
public static void main(String[] args) {
ZigzagConvert zz = new ZigzagConvert();
int nRows = 2;
String s = zz.convert("ABCDE", nRows);
System.out.println(s);
}
}
参考：