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"
解题思路：
观察题目，靠眼力寻找规律即可。如果没有读懂ＺｉｇＺａｇ的话，请移步
http://blog.csdn.net/zhouworld16/article/details/14121477
java实现：
static public String convert(String s, int nRows) {
if (s == null || s.length() <= nRows || nRows <= 1)
return s;
StringBuffer sb = new StringBuffer();
for (int i = 0; i < s.length(); i += (nRows - 1) * 2)
sb.append(s.charAt(i));
for (int i = 1; i < nRows - 1; i++) {
for (int j = i; j < s.length(); j += (nRows - 1) * 2) {
sb.append(s.charAt(j));
if (j + (nRows - i - 1) * 2 < s.length()) {
sb.append(s.charAt(j + (nRows - i - 1) * 2));
}
}
}
for (int i = nRows - 1; i < s.length(); i += (nRows - 1) * 2)
sb.append(s.charAt(i));
return new String(sb);
}
C++实现如下：
#include<string>
using namespace std;
class Solution {
public:
string convert(string s, int numRows) {
if (s.length() <= numRows || numRows <= )
return s; string sb;
for (int i = ; i < s.length(); i += (numRows - ) * )
sb+=s[i];
for (int i = ; i < numRows - ; i++) {
for (int j = i; j < s.length(); j += (numRows - ) * ) {
sb+=s[j];
if (j + (numRows - i - ) * < s.length())
sb+=s[j + (numRows - i - ) * ];
}
} for (int i = numRows - ; i < s.length(); i += (numRows - ) * )
sb += s[i];
return sb;
}
};