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]!=''){
	        			 sb.append(String.valueOf(zigzag[m][n]));
	        		}
	        	}
	       
	        return sb.toString();
	    }
}




原文地址:https://www.cnblogs.com/wennian/p/5036917.html