【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;

     }

 };

巴特西

【JAVA、C++】LeetCode 006 ZigZag Conversion

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