ZigZag Conversion
The string "PAYPALISHIRING" is written in 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 N
A P L S I I G
Y I R
And then read line by line: "PAHNAPLSIIGYIR"
Write the code that will take a string and make this conversion given a number of rows:
string convert(string text, int nRows);
convert("PAYPALISHIRING", 3) should
return "PAHNAPLSIIGYIR".
思路：列输入-->行输出，将zigzag字符串看成是字符数组，为空的位置其值为null
先确定数组的列数（nRows即为数组的行数）
time=440ms
accepted
public class Solution {
public String convert(String s, int nRows) {
int length=s.length();
if(length<=nRows||nRows==1)
return s;
int updown=0,i=0,j=1,count=0,nLines=0;
while(count<length){
if(updown==0){
if(i>nRows-1){
updown=nRows-1;
i=nRows-2;
}else{
count++;
i++;
}
}
if(updown==nRows-1){
if(i<1){
updown=0;
i=0;
j++;
}else{
count++;
i--;
j++;
}
}
}
nLines=j;
System.out.println("j="+j);
char[][] zigzag=new char[nRows][nLines];
updown=0;i=0;j=0;count=0;
while(count<length){
if(updown==0){
if(i>nRows-1){
updown=nRows-1;
i=nRows-2;
j++;
}else{
zigzag[i][j]=s.charAt(count);
count++;
i++;
}
}
if(updown==nRows-1){
if(i<1){
updown=0;
i=0;
}else{
zigzag[i][j]=s.charAt(count);
count++;
i--;
j++;
}
}
}
StringBuffer sb = new StringBuffer();
for(int m=0;m<nRows;m++)
for(int n=0;n<nLines;n++){
if(zigzag[m][n]!='\0'){
sb.append(String.valueOf(zigzag[m][n]));
}
}
return sb.toString();
}
}