【JAVA、C++】LeetCode 006 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"

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;
     }
 };