[Leetcode] [Java] ZigZag Conversion

Topic:

The string is "PAYPALISHIRING" written with 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   NA p L S i i GY   i   R

And then read on line : "PAHNAPLSIIGYIR"

Write the code that would take a string and make this conversion given a number of rows:

String convert (string text, int nRows);

convert("PAYPALISHIRING", 3)should return "PAHNAPLSIIGYIR".
Test instructions

Rearranges the string by column in the given format, followed by rows to return the new string.

Algorithm Analysis: We take the character in the string ordinal to arrange the analysis

* One: 2 rows, 1 to N of the sort

* 1 3 5 7 9 ...
* 2 4 6 8 10 ...

* Two: 3 Rows of time, 1 to n sort of

* 1 5 9 ...
* 2 4 6 8 10 ...
* 3 7 11 ...

* Three: 4 rows of time, 1 to n sort of

* 1 7 13 ...
* 2 6 8 12 14 ...
* 3 5 9 11 15 ...
* 4 10 16 ...

* Did you find the rules? If not found, you can continue to write 5 rows of case. Soon you'll be able to find patterns. This is a way to solve the problem.

* When we have a difficult problem, let's consider a simple situation and see if we can find a pattern. This topic, we write to these special cases.

* We find the following rules, here we assume we are divided into M-rows:

* 1 row I starting from I

* 2 the interval of two digits of row i is 2 (i-1), 2 (m-i) alternating

The code is as follows:

public class Solution {public    string convert (string s, int numrows)     {        string result = "";                if (numrows = = 1)         {          return s;        }                for (int i = 0; i < numrows; i++)         {          int j = i;          Boolean flag = true;//Toggles between two intervals via flag          while (J < S.length ())           {            result+= (S.charat (j));                        if (i = = 0 | | i = = numRows-1)//Start and end lines are spaced between two numbers is 2 (i-1)              J + = 2 * (numRows-1);            else             {              if (flag)              {                J + = 2 * (numRows-1-i);                Flag = false;              }              else               {                J + = 2 * i;                Flag = true;        }}} return result;}    }

Copyright NOTICE: This article is the original article of Bo Master, reprint annotated source

[Leetcode][java] ZigZag Conversion