LeetCode6-ZigZag convert

【题目】

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"

.
将字符串锯齿状排列后获取其横排重构后的字符串；

【思路】

找规律，发现锯齿状的每一个元素与其所在的行列的关系如下：

0		2r-2		2(2r-2)	……
1	(2r-2)-1	(2r-2)+1	(2*(2r-2))-1	(2*(2r-2))+1	……
2	(2r-2)-2	(2r-2)+2	(2*(2r-2))-2	(2*(2r-2))+2	……
3	(2r-2)-3	(2r-2)+3	(2*(2r-2))-3	(2*(2r-2))+3	……
……	……	……	……	……	……
r-1	(2r-2)-(r-1)	(2r-2)+(r-1)	(2*(2r-2))-(r-1)	(2*(2r-2))+(r-1)	……

当numRows=1时，排序后的字符串和排序前的相同
【Java代码】

public class Solution_6_zigzag_convert {
public String convert(String s, int numRows){
String result = "";
if(numRows == 1){
result = s;
return result;
}
for(int i = 0 ; i < numRows&&i < s.length() ; i++){
result += s.charAt(i);
int j = 1;
if(i == 0)
while(j*(2*numRows-2)<s.length())
result += s.charAt(j++*(2*numRows-2));
else if(i == numRows - 1)
while(j*(2*numRows-2)+numRows-1<s.length())
result += s.charAt(j++*(2*numRows-2)+numRows-1);
else{
while(j*(2*numRows-2)-i<s.length()){
result += s.charAt(j*(2*numRows-2)-i);
if(j*(2*numRows-2)+i<s.length())
result += s.charAt(j*(2*numRows-2)+i);
j++;
}
}
}
return result;
}
}