Leetcode 6 ZigZag Conversion (z-type conversion)

Translation

字符串“PAYPALISHIRING”通过一个给定的行数写成如下这种Z型模式：P   A   H   NII GY   I   R然后一行一行的读取：“PAHNAPLSIIGYIR”写代码读入一个字符串并通过给定的行数做这个转换：string convert(string text, int nRows);调用convert("PAYPALISHIRING"3)，应该返回"PAHNAPLSIIGYIR"。

Original

Thestring "Paypalishiring"  isWritteninchA zigzag pattern onAgiven  Number  ofRows like this: (Want toDisplay this patterninchA fixed font forBetter legibility) P A H NA p L S i i GY i RAnd Then ReadLine byLine"Pahnaplsiigyir"Write theCode thatwould take astring  andMake this conversiongivenA Number  ofRowsstringConvertstring text, int nRows); CONVERT ("Paypalishiring",3) shouldreturn "Pahnaplsiigyir".

If you still do not understand the meaning of the topic, see ...

 Public classsolution{ Public string Convert(stringSintNumRows) {if(NumRows = =1)returnS StringBuilder Strbuilder =NewStringBuilder ();intLengthofgroup =2* NumRows-2;//As shown, the length of each group is 4         for(introw =0; Row < NumRows; row++)//Follow the sequence of rows from line No. 0 to NumRows-1{if(Row = =0|| row = = NumRows-1)//Here is responsible for line No. 0 and NumRows-1{ for(intj = row; J < S.length;                J + = Lengthofgroup) {strbuilder.append (s[j]); }            }Else                   //Here is responsible for all rows between line No. 0 and NumRows-1{intCurrentRow = row;//Move right in the current line (see)                BOOLFlag =true;intChildLenOfGroup1 =2* (NumRows-1-row);//How to say it ... Each index of the middle row                intChildLenOfGroup2 = Lengthofgroup-childlenofgroup1; while(CurrentRow < S.length) {strbuilder.append (S[currentrow]);if(flag) CurrentRow + = ChildLenOfGroup1;ElseCurrentRow + = childLenOfGroup2;                flag =!flag; }            }        }returnStrbuilder.tostring (); }}

The code for C + + must be there:

Class Solution { Public:string Convert(stringSintNumRows) {if(numrows==1)returnSstringStr="";intlengthofgroup=2*numrows-2; for(introw=0; row<numrows;row++) {if(row==0|| row==numrows-1){ for(intCurrentrow=row;currentrow<s.length (); currentrow+=lengthofgroup) {Str+=s[currentrow]; }            }Else{intCurrentrow=row;BOOLflag=true;intchildlenofgroup1=2* (numrows-1-row);intChildlenofgroup2=lengthofgroup-childlenofgroup1; while(Currentrow<s.length ()) {Str+=s[currentrow];if(flag) Currentrow+=childlenofgroup1;Elsecurrentrow+=childlenofgroup2;                Flag=!flag; }            }        }returnStr }};

As for Java, of course there are ... But every time I just copied the code of C # and changed it.

 Public classSolution { PublicStringConvert(String S,intNumRows) {if(NumRows = =1)returnS StringBuilder Strbuilder =NewStringBuilder ();intLengthofgroup =2* NumRows-2; for(introw=0; Row < NumRows; row++) {if(Row = =0|| row = = NumRows-1){ for(intCurrentRow = row; CurrentRow < S.length ();                  CurrentRow + = Lengthofgroup) {strbuilder.append (S.charat (CurrentRow)); }              }Else{intCurrentRow = row; Boolean flag =true;intChildLenOfGroup1 =2* (NumRows-1-row);intChildLenOfGroup2 = Lengthofgroup-childlenofgroup1; while(CurrentRow <s.length ()) {Strbuilder.append (S.charat (CurrentRow));if(flag) CurrentRow + = ChildLenOfGroup1;ElseCurrentRow + = childLenOfGroup2;                  flag =!flag; }              }          }returnStrbuilder.tostring (); }}

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Leetcode 6 ZigZag Conversion (z-type conversion)