题目描述：
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 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 s, int numRows);
Example 1:
Input: s = "PAYPALISHIRING", numRows = 3
Output: "PAHNAPLSIIGYIR"
Explanation:
P A H N
A P L S I I G
Y I R
Example 2:
Input: s = "PAYPALISHIRING", numRows = 4
Output: "PINALSIGYAHRPI"
Explanation:
P I N
A L S I G
Y A H R
P I
My Solution（19ms，38.4MB）
首先想到的办法，有点绕，用一个二维数组按照之字形储存起来，然后逐行逐列地去读取，然后拼接成一个字符串。
class Solution {
public String convert(String s, int numRows) {
if(numRows<2){return s;}//防止除数为零
int index = 0,row=numRows-2;
//columns实际上是指二维数组列下表的最大值
int columns = s.length()/(2*numRows-2)*(numRows-1);
int remainder = s.length()%(2*numRows-2);
if(remainder>numRows){columns = remainder - numRows + columns;}
String[][] zigzag = new String[numRows][columns+1];
//开始逐行填充数组
for(int j = 0;j<columns+1;j++){
if(j%(numRows-1)==0){//竖笔
for(int i = 0;i<numRows;i++){
if(index<s.length()){
zigzag[i][j] = s.substring(index,index+1);index++;
row = numRows-2;
}
}
}else{
if(index<s.length()){
zigzag[row][j] = s.substring(index,index+1);index++;
row--;
}
}
}
//开始拼接结果字符串
String result = "";
for(int i =0 ; i<numRows ;i++){
for(int j=0;j<columns+1;j++){
if(zigzag[i][j]!=null){
result = result + zigzag[i][j];
}
}
}
return result;
}
}Visit By Row（3ms，36MB）
发现数字间的规律，直接逐行去读取（效率最高）
class Solution {
public String convert(String s, int numRows) {
if (numRows == 1) return s;
StringBuilder ret = new StringBuilder();
int n = s.length();
int cycleLen = 2 * numRows - 2;
//逐行去读取
for (int i = 0; i < numRows; i++) {
for (int j = 0; j + i < n; j += cycleLen) {
ret.append(s.charAt(j + i));
//排除第一行和最后一行，因为两竖笔之间没有数字
if (i != 0 && i != numRows - 1 && j + cycleLen - i < n)
ret.append(s.charAt(j + cycleLen - i));
}
}
return ret.toString();
}
}