Zigzag transformation JAVA

Date: 2020-10-9

Author: 19th LZ

Tag: JAVA

Title description

Arrange a given character string in a zigzag pattern from top to bottom and from left to right according to a given number of lines.

For example, when the input string is "LEETCODEISHIRING" and the number of rows is 3, the arrangement is as follows:

L   C   I   R
E T O E S I I G
E   D   H   N

After that, your output needs to be read line by line from left to right to generate a new string, such as: "LCIRETOESIIGEDHN".

Please implement this function that transforms a string into a specified number of lines:

string convert(string s, int numRows);

Example:

示例 1:

输入: s = "LEETCODEISHIRING", numRows = 3
输出: "LCIRETOESIIGEDHN"

示例 2:

输入: s = "LEETCODEISHIRING", numRows = 4
输出: "LDREOEIIECIHNTSG"
解释:

L     D     R
E   O E   I I
E C   I H   N
T     S     G

Problem-solving ideas:

Use 2numRows-2 as a loop cyc, and
just output it directly according to the formula. The
first line and the last line have a difference of one cyc between each character;
each character in the middle part has a difference of cyc-2i (i means the first row) , And 2i;

Code

class Solution {
    public String convert(String s, int numRows) {
      
      if(numRows==1)
         return s.toString();

      StringBuilder result= new StringBuilder();

      int all=s.length();
      int cyc=2*numRows-2;
        
      for(int j=0;j<all;j+=cyc){                      //第一行
          result.append(s.charAt(j));
      }

      for(int i=1;i<numRows-1;i++){                  //中间部分
          for(int j=i;j<all;){
            
            result.append(s.charAt(j));
            j+=cyc-2*i;

            if(j<all){
            result.append(s.charAt(j));
            j+=2*i;
            }
          }
      }

      for(int j=numRows-1;j<all;j+=cyc){             //最后一行
          result.append(s.charAt(j));
      }
     return result.toString();
    }
}