leetcode6-ZigZag Conversion

最新推荐文章于 2023-03-04 23:54:09 发布

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本文介绍了一种将字符串以Z字形方式写入指定行数再逐行读取的算法实现。通过C++代码示例展示了如何将输入字符串转换为所需的Z字形布局并返回转换后的结果。

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".

class Solution {
public:
    string convert(string s, int nRows) {
		int nLength=s.length()-1;
		string s1(s.length(),sizeof(char));
		int nIndRow,nIndCol;
		int nIndS1=0;
		if(nRows==1)
			return s;
		for(nIndRow=0;nIndRow<=nRows-1;nIndRow++)
		{	
			for(nIndCol=0;nIndCol<=nLength/(2*nRows-2);nIndCol++)
			{
				if(nIndRow==0)
				{
					s1[nIndS1]=s[nIndCol*(2*nRows-2)];
					nIndS1++;
				}
				if(nIndRow==nRows-1)
				{
					if(nIndCol!=nLength/(2*nRows-2)||nIndCol*(2*nRows-2)+nRows-1<=nLength)
					{
						s1[nIndS1]=s[nIndCol*(2*nRows-2)+nRows-1];
						nIndS1++;
					}
				}
				if(nIndRow!=0&&nIndRow!=nRows-1)
				{
					if(nIndCol==nLength/(2*nRows-2))
					{
						int nLa=nLength%(2*nRows-2);
						if(nLa>=nIndRow)
						{
							s1[nIndS1]=s[nIndCol*(2*nRows-2)+nIndRow];
							nIndS1++;
						}
						if(2*nRows-2-nLa<=nIndRow)
						{
							s1[nIndS1]=s[nIndCol*(2*nRows-2)+2*(nRows-1)-nIndRow];
							nIndS1++;
						}
					}
					else
					{
						s1[nIndS1]=s[nIndCol*(2*nRows-2)+nIndRow];
						nIndS1++;
						s1[nIndS1]=s[nIndCol*(2*nRows-2)+2*(nRows-1)-nIndRow];
						nIndS1++;
					}
				}
			}
		}
		return s1;
    }
};