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

 1 public class Solution {

 2     public String convert(String s, int nRows) {

 3         if (s == null) {

 4             return null;

 5         }

 6         

 7         // 第一个小部分的大小        

 8         int size = 2 * nRows - 2;

 9         

10         // 当行数为1的时候，不需要折叠。

11         if (nRows <= 1) {

12             return s;

13         }

14         

15         StringBuilder ret = new StringBuilder();

16         

17         int len = s.length();

18         for (int i = 0; i < nRows; i++) {

19             // j代表第几个BLOCK

20             for (int j = i; j < len; j += size) {

21                 ret.append(s.charAt(j));

22                 

23                 // 即不是第一行，也不是最后一行，还需要加上中间的节点

24                 int mid = j + size - i * 2;

25                 if (i != 0 && i != nRows - 1 && mid < len) {

26                     char c = s.charAt(mid);

27                     ret.append(c);

28                 }

29             }

30         }

31         

32         return ret.toString();

33     }

34 }

View Code

 2015.1.4 redo:

 1 public class Solution {

 2     public String convert(String s, int nRows) {

 3         if (s == null) {

 4             return null;

 5         }

 6         

 7         // corner case;

 8         if (nRows == 1) {

 9             return s;

10         }

11         

12         // The number of elements in a section.

13         int section = 2 * nRows - 2;

14         StringBuilder sb = new StringBuilder();

15         

16         int len = s.length();

17         for (int i = 0; i < nRows; i++) {

18             for (int j = i; j < len; j += section) {

19                 char c = s.charAt(j);

20                 sb.append(c);

21                 

22                 // The middle rows.

23                 int mid = j + section - 2 * i;

24                 // bug 2: the mid is out of range.

25                 if (i != 0 && i != nRows - 1 && mid < len) {

26                     // bug 1: forget a ')'

27                     sb.append(s.charAt(mid));

28                 }

29             }

30         }

31         

32         return sb.toString();

33     }

34 }

View Code

 

请至主页君的GIT HUB:

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