leetcode第六题 ZigZag Conversion （java）

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();
	    }
}