ZigZag Conversion

题目原型：

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

Write the code that will take a string and make this conversion given a number of rows:

string convert(string text, int nRows);

基本思路：

这道题我自己想了一种思路，完全是按照矩阵的路径来做的，虽说做出来了，但是比较笨重。

	public String convert(String s, int nRows)
	{
		if(nRows==1)
		    return s;
		if(nRows==2)
		{
		    String s1 = "";
		    String s2 = "";
		    for(int index = 0;index<s.length();index++)
		    {
		        if(index%2==0)
		            s1+=s.charAt(index);
		        else
		            s2+=s.charAt(index);
		    }
		    return s1+s2;
		}
		StringBuffer strbuf = new StringBuffer();
		int row = 0;
		int column = 0;
		row = nRows;
		column = (nRows-1)*(s.length()/(nRows+(nRows-2)));
		
		if(s.length()%(nRows+(nRows-2))<=nRows)
			column+=1;
		else
			column=column+(s.length()%(nRows+(nRows-2))-nRows)+1;
		Character[][] ch = new Character[row][column];
		
		int i = 0;//表示行
		int j = 0;//表示列
		
		int index = -1;//遍历字符串要用
		
		int count = 0;//计数步数
		for(;index<s.length()-1;)
		{
			//先向下走nRows步
			count = nRows;
			while(count>0&&index<s.length()-1)
			{
				index++;
				ch[i][j] = s.charAt(index);
				i++;
				count--;
			}
			i--;//向上退回一步
			//斜向上走nRows-2步
			count = nRows - 2;
			while(count>0&&index<s.length()-1)
			{
				index++;
				i--;
				j++;
				ch[i][j] = s.charAt(index);
				count--;
			}
			i--;
			j++;//斜向上走到待定位置
		}
		
		for(i=0;i<row;i++)
		{
			for(j=0;j<column;j++)
			{
				if(ch[i][j]!=null)
					strbuf.append(ch[i][j]);
			}
		}
		return strbuf.toString();
	}

	public String convert(String s, int nRows)
	{
		if (s == null || s.length() <= nRows || nRows <= 1)
			return s;
		StringBuffer sb = new StringBuffer();
		// the first row
		for (int i = 0; i < s.length(); i += (nRows - 1) * 2)
		{
			sb.append(s.charAt(i));
		}

		for (int i = 1; i < nRows - 1; i++)
		{
			for (int j = i; j < s.length(); j += (nRows - 1) * 2)
			{
				sb.append(s.charAt(j));
				if (j + (nRows - i - 1) * 2 < s.length())
				{
					sb.append(s.charAt(j + (nRows - i - 1) * 2));
				}
			}
		}
		// the last row
		for (int i = nRows - 1; i < s.length(); i += (nRows - 1) * 2)
		{
			sb.append(s.charAt(i));
		}
		return sb.toString();
	}

ZigZag Conversion,布布扣,bubuko.com

ZigZag Conversion

原文：http://blog.csdn.net/cow__sky/article/details/20932175