ZigZag Conversion

最新推荐文章于 2025-05-22 17:48:42 发布

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本文介绍了一种将字符串以Z字形方式排列并按行读取的算法实现。该算法接受一个字符串和行数作为输入参数，返回转换后的字符串。通过实例演示了如何使用这一方法，并提供了详细的代码实现。

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".

我写的这段代码实在太复杂了，做完后网上一搜大多人写的，思路比我简单多了

class Solution {
public:
	string convert(string s, int nRows)
	{
		string res;
		size_t len = s.length();
		if (0 == len || nRows < 1) return res;
		if (1 == nRows || 1 == len) return s;

		int tmp = len/(2*nRows - 2);
		int range = len%(2*nRows - 2) == 0 ? tmp-1 : tmp;
		for (int i = 0; i < nRows; i++)
		{
			for (int j = 0; j <= range; j++)
			{
				if (0 == i && j*2*(nRows - 1) < len)
					res.push_back(s[j*2*(nRows - 1)]);
				else if (i + 1 == nRows && (j*2 + 1)*(nRows - 1) < len)
					res.push_back(s[(j*2 + 1)*(nRows - 1)]);
				else
				{
					if (j*2*(nRows - 1) + i < len)
						res.push_back(s[j*2*(nRows - 1) + i]);
					if ((j*2 + 1)*(nRows - 1) + nRows - 1 - i < len)
						res.push_back(s[(j*2 + 1)*(nRows - 1) + nRows - 1 - i]);
				}
			}
		}
		return res;
	}
};