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形方式排列并按行读取的算法实现。通过分析不同行数对字符串排列的影响，提供了Java代码示例，展示了如何根据指定的行数将输入字符串转换为Zigzag模式。

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".

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package leetcode;

/**
 * 很容易看出外层的循环周期；
 * 若s="0123456789",当numRows=3时,打印出 048/159/26
 *                 当numRows=4时，打印出06/17/28/39
 * 故循环周期是cyc=2*numRows-2；
 * 对于首行和末行之间的行，期间还有一个周期,这一次距离本周期起始字符的距离是 2*nRows-2-2*i
 * @author Mouse
 *
 */

public class Solution {
	public static String convert(String s, int numRows) {
		// 获得s字符串的长度
		int len = s.length();
		if (len<=2||numRows==1) {
			return s;
			
		}
		String str="";
		int cyc=2*numRows-2;//循环周期
		for (int i = 0; i < numRows; i++) {
			for (int j = i; j <len; j+=cyc) {
				str+=s.charAt(j);	
				if (i>0&&i<numRows-1) {//非首行和末行时还要加一个
					int t=j+cyc-2*i;
					if (t<len) {
						str+=s.charAt(t);
					}
			}
			
		}
		}
		
		return str;

	}

	public static void main(String[] args) {
		convert("0123456789", 3);

	}

}