Leetcode # ZigZag Conversion #

My Python solution:

"" "Programmer:eofdate:2015.04.14file:zigzag.pye-mail: [email protected]" "" Class so         Lution:def convert (self, S, nRows): length = Len (s) ret_string = "" i = 0 row = 0                Counter = 0 if nRows! = 1:while Row < nrows:i = Row col = 0 While I < length and counter < Length:ret_string + S[i] if row = = 0                        or row = = NRows-1: i + = (nRows-1) * elif Col% 2 = = 0:                    i + = (nrows-row-1) * else:i + = row*2 Col + + 1 Counter + = 1 row + 1 Else:return s return ret_string#-----------Ju St for testing----------S = solution () print S.convert ("AB", 2) Print S.convert ("paypalishiring", 3) print S.convert (" Paypalishiring ", 4)

The Java answer to the Triumph charge:

Package Zigzag_conversion;public class Solution {/*0 [Totalgap] 6                           C1 [Totalgap-2row] 5 [2row] 7 B D2 4                                         8 a E3 9 F*/public string Convert (string s, int nRows) {if (nRows = = 1) {return s;} int len = S.length (); StringBuilder builder = new StringBuilder (len), int totalgap = (nRows-1) << 1;for (int i = 0; i < len; i + = Tota LGAP) {builder.append (S.charat (i));} for (int row = 1; row < nRows-1; row++) {for (int i = row, gap = row << 1; i < len; gap = Totalgap-gap, I + = Gap) {builder.append (S.charat (i));}} for (int i = nRows-1; i < len; i + = Totalgap) {builder.append (S.charat (i));} return builder.tostring ();} public static void Main (string[] args) {System.out.println (New solution (). CONVERT ("ABCD", 4));}} 

The C + + solution of our Hao God:

source:https://oj.leetcode.com/problems/zigzag-conversion///Author:hao chen//date:2014-07-17/**************** * * The string "Paypalishiring" is written in a Zigzag pattern on a given number of rows type this: * (the want to display the this pattern in a fixed font for better leg  ibility) * * 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 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". * * *************************************************** /#include <iostream> #include <vector> #include <string>using Namespace Std;string Convert (string s, int nRows) {//the cases no need to do anything if (Nrows<=1 | | nrows>=         S.size ()) return s; Vector<string> R (nRows);    int row = 0;    int step = 1;        for (int i=0; i<s.size (); i + +) {if (row = = nRows-1) Step =-1;        if (row = = 0) step = 1;        cout << Row <<endl;        R[row] + = s[i];    Row + = step;    The string result;    for (int i=0; i<nrows; i++) {result + = R[i]; } return result;    int main (int argc, CHAR**ARGV) {string S;    int R;    s = "paypalishiring";    R = 3; cout << s << ":" << convert (S, 3) << Endl;}

Leetcode # ZigZag Conversion #