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：：

The key here is to figure out the index pattern, just draw several concrete example could hele:

nRow = 2:



0 2 4 6 8 10 12

1 3 5 7 9 11 13



nRow = 3: 



0     4      8     12 

1  3  5   7  9  11

2     6     10



nRow = 4:

0      6       12

1   5  7    11 13

2 4    8  10

3      9

So inorder to get the final string, we need to scan from the left to right row by row, for the first and last row, the difference
between every two is 2 * nRow – 2, and for the middle say i-th rows, the difference between every two is either 2 * nRow – 2 – 2 * i
or 2 * i in turn. Following this, a linear scan of the original string could give us the final result string by pushing the corresponding character  at specific index into the final resulted string.

第0, len-1行各个字符之间的间隔为2×nRows - 2

第1~len-2行之间会有两个固定间隔之间会有其他字符插入

规律：index + 2 * nRows - 2 * i - 2, 这里i是指行号

 

 1 public class Solution {

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

 3         // Start typing your Java solution below

 4         // DO NOT write main() function

 5         if (s == null || s.length() == 0 || nRows <= 1) {

 6             return s;

 7         }

 8         int len = s.length();

 9         StringBuffer buffer = new StringBuffer();

10         int diff = 2 * nRows - 2;

11         for (int i = 0; i < nRows && i < len; i++) {

12             if (i == 0 || i == nRows - 1) {

13                 buffer.append(s.charAt(i));

14                 int index = i;

15                 while (index + diff < len) {

16                     buffer.append(s.charAt(index + diff));

17                     index = index + diff;

18                 }

19             } 

20             else{

21                 buffer.append(s.charAt(i));

22                 int index = i;

23                 while (2 * nRows - 2 * i - 2 + index < len

24                         || index + diff < len) {

25                     if(2 * nRows - 2 * i - 2 + index < len){

26                         buffer.append(s.charAt(2 * nRows - 2 * i - 2 + index));

27                     }

28                     if (index + diff < len) {

29                         buffer.append(s.charAt(index + diff));

30                     }

31                     index += diff;

32                 }

33             }

34         }

35         return buffer.toString();

36     }

37 }

 第二遍：

 1 public class Solution {

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

 3         // Note: The Solution object is instantiated only once and is reused by each test case.

 4         StringBuffer sb = new StringBuffer();

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

 6             int j = i;

 7             int k = 0;

 8             while(j < s.length()){

 9                 sb.append(s.charAt(j));

10                 if ( nRows == 1 ){

11                     j ++;

12                 }

13                 if(i == 0 || i == nRows - 1){

14                     j += (nRows - 1) * 2;

15                 }else{

16                     j += (k % 2 == 1) ? ((nRows - 1) * 2- nRows * 2 + (i + 1) * 2) : (nRows * 2-(i + 1) * 2);

17                 }

18                 k ++;

19             }

20         }

21         return sb.toString();

22     }

23 }

 第三遍：

 1 public class Solution {

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

 3         if(s == null || s.length() == 0) return "";

 4         if(nRows == 1) return s;

 5         StringBuilder sb = new StringBuilder();

 6         for(int i = 0; 2*(nRows - 1) * i < s.length(); i ++){//first row

 7             sb.append(s.charAt(2*(nRows - 1) * i));

 8         }

 9 

10         for(int j = 1; j < nRows - 1; j ++){

11             for(int i = 0; (j + 2 * i * (nRows - 1)) < s.length(); i ++){

12                 sb.append(s.charAt((j + 2 * i * (nRows - 1))));

13                 if((2 * (nRows - 1) * (i + 1) - j) < s.length())

14                     sb.append(s.charAt(2 * (nRows - 1) * (i + 1) - j));

15             }

16         }

17         

18         for(int i = 0; (nRows - 1) * (2 * i + 1) < s.length(); i ++){//last row

19             sb.append(s.charAt((nRows - 1) * (2 * i + 1)));

20         }

21         return sb.toString();

22     }

23 }