"Leetcode-Interview algorithm classic-java Implementation" "006-zigzag Conversion (z-font conversion)"

"006-zigzag Conversion (z-font conversion)" "leetcode-Interview algorithm classic-java Implementation" "All topics Directory Index" Original Question

The string "Paypalishiring" is written with a zigzag pattern on a given number of rows like this: (You could want to display T His pattern in a fixed font for better legibility)
P A H N
Aplsiig
Y I R
And then read 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".

Main Topic

Enter a string and the specified number of lines to output the characters in Z-font.

Thinking of solving problems

Calculates the maximum number of columns of a character, creates a one-dimensional array based on the number of columns and rows, computes the position of each character in a Fu Zheng-dimensional array, and then makes a compact operation of the characters in a one-dimensional array, returning the result.

Code Implementation

 Public  class solution {     PublicString Convert (String s,intNRows) {if(s = =NULL|| S.length () <= NRows | | NRows = =1) {returnS }int Index= S.length ();intRowlength =0;//Calculates the length of the line, including the last line break character        intSlash = NRows-2;//A slash to remove the number of rows occupied by the end and end         while(Index>0) {//A column in a vertical shape            Index-= NRows; rowlength++;//Diagonal Number of columns             for(inti =0; I < slash &&Index>0; i++) {rowlength++;Index--; }        }Char[] result =New Char[NRows * rowlength];//Save an array of results, and the last column to save the line break         for(inti =0; i < result.length; i++) {//Initialize to a spaceResult[i] ="'; }intCurcolumn =0;//number of rows currently being processed        Index=0; while(Index< S.length ()) {//Handle Vertical line             for(inti =0; I < nRows &&Index< S.length (); i++) {Result[rowlength * i + curcolumn] = S.charat (Index);Index++; } curcolumn++;//Handle slash             for(inti = nRows-2; i >0&&Index< S.length (); i--) {Result[rowlength * i + curcolumn] = S.charat (Index); curcolumn++;Index++; }        }//System.out.println (new String (result));        //Compact manipulation of character arrays        Index=0; while(Index< S.length () && result[Index] !="') {//Find the first character position that is a space            Index++; }intNext =Index+1; while(Index< S.length ()) { while(Next < Result.length && Result[next] = ="') {//Find elements that are not spacesnext++; } result[Index] = Result[next];Index++;        next++;        } System.out.println (s); System.out.println (NewString (Result,0,Index));return NewString (Result,0,Index); }}

Evaluation Results

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"Leetcode-Interview algorithm classic-java Implementation" "006-zigzag Conversion (z-font conversion)"