week_10_ ZigZag Conversion

最新推荐文章于 2022-08-30 20:42:29 发布

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本文介绍了一种将字符串以ZigZag形式分布到多行再按行读取的算法实现，通过模拟规则在二维数组中填充字符，最终得到转换后的字符串。

Description

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".

Solution

这种题目没什么算法可言，按照ZigZag给定的格式在二维数组中相应的填充字符即可。

代码实现如下：

class Solution {
public:
	// 模拟规则在数组中进行填充
	string convert(string s, int numRows) {
        // special case
        if (numRows == 0 || numRows == 1)
            return s;
		// 申请数组并初始化
		int rows = numRows;
		int columns = s.length();
		ZigZag = new char*[rows];
		for (int i = 0; i < rows; i++) {
			ZigZag[i] = new char[columns];
			for (int j = 0; j < columns; j++)
				ZigZag[i][j] = '\0';
		}
		// 填充
		index = row = column = 0;
		while (index < s.length()) {
			while (index < s.length() && row < rows - 1) {
				ZigZag[row][column] = s[index++];
				row++;
			}
			while (index < s.length() && row > 0) {
				ZigZag[row][column] = s[index++];
				row--;
				column++;
			}
		}
		// 输出、释放内存
		string result;
		for (int i = 0; i < rows; i++) {
			for (int j = 0; j < columns; j++)
				if (ZigZag[i][j] != '\0')
					result += ZigZag[i][j];
			delete[] ZigZag[i];
		}
		delete[] ZigZag;
		// 返回结果
		return result;
	}
private:
	char** ZigZag = NULL;
	int index;
	int row;
	int column;
};

在LeetCode提交结果如下：