LeetCode: ZigZag Conversion 解题报告

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"
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".

SOLUTION 1:

使用以下算法会比较简单。

两个规律：

1 两个zigzag之间间距为2*nRows-2

2 每个zigzag中间（在j和j+interval之间）位置为j+interval-2*i

注意：当Rows = 1时，此方法不适用，因为size = 0,会造成死循环。所以Rows = 1时，需要独立处理。

引自：http://blog.csdn.net/fightforyourdream/article/details/16881517

 public class Solution {
     public String convert(String s, int nRows) {
         if (s == null) {
             return null;
         }

         // 第一个小部分的大小
         int size =  * nRows - ;

         // 当行数为1的时候，不需要折叠。
         if (nRows <= ) {
             return s;
         }

         StringBuilder ret = new StringBuilder();

         int len = s.length();
         for (int i = ; i < nRows; i++) {
             // j代表第几个BLOCK
             for (int j = i; j < len; j += size) {
                 ret.append(s.charAt(j));

                 // 即不是第一行，也不是最后一行，还需要加上中间的节点
                 int mid = j + size - i * ;
                 if (i !=  && i != nRows -  && mid < len) {
                     char c = s.charAt(mid);
                     ret.append(c);
                 }
             }
         }

         return ret.toString();
     }
 }

2015.1.4 redo:

 public class Solution {
     public String convert(String s, int nRows) {
         if (s == null) {
             return null;
         }

         // corner case;
         if (nRows == 1) {
             return s;
         }

         // The number of elements in a section.
         int section = 2 * nRows - 2;
         StringBuilder sb = new StringBuilder();

         int len = s.length();
         for (int i = 0; i < nRows; i++) {
             for (int j = i; j < len; j += section) {
                 char c = s.charAt(j);
                 sb.append(c);

                 // The middle rows.
                 int mid = j + section - 2 * i;
                 // bug 2: the mid is out of range.
                 if (i != 0 && i != nRows - 1 && mid < len) {
                     // bug 1: forget a ')'
                     sb.append(s.charAt(mid));
                 }
             }
         }

         return sb.toString();
     }
 }

请至主页君的GIT HUB:

https://github.com/yuzhangcmu/LeetCode_algorithm/blob/master/string/Convert.java