leetcode ZigZag Conversion 006

最新推荐文章于 2021-11-05 22:58:08 发布

walter1990

最新推荐文章于 2021-11-05 22:58:08 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此方法对特定字符串进行转换。

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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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#include <iostream>
#include <vector>
#include <string>
using namespace std;

class Solution {
	public:
		string convert(string s, int numRows) {
			string ans="";
			if(s=="" || numRows==1) return s;
			int len = numRows+numRows-2;
			int num = s.length()/len, remain=s.length()-num*len;

			for(int i=0; i < num; i++) {
				ans+=s[i*len];
			}

			if(remain>0) ans+=s[num*len];

			for (int j = 1; j < numRows-1; ++j)
			{
				for (int i = 0; i < num; ++i)
				{
					ans += s[i*len+j];
					ans += s[i*len+j+2*(numRows-1-j)];
				}
				if(remain > 0) {
					if(num*len+j < s.length())
						ans += s[num*len+j];
					if(num*len+j+2*(numRows-1-j) < s.length())
						ans+= s[num*len+j+2*(numRows-1-j)];
				}
			}

			for (int i = 0; i < num; ++i)
			{
				ans+=s[i*len+numRows-1];
			}

			if(remain > 0) {
				if(num*len+numRows-1 < s.length())
					ans+= s[num*len+numRows-1];
			}

			return ans;
		}
};