【LeetCode】ZigZag Conversion 题解

最新推荐文章于 2020-12-12 15:35:57 发布

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本文详细介绍了zigzag模式字符串转换算法的实现方法，包括C++和Python版本的代码示例，通过解析输入字符串并按照指定行数进行zigzag排列，最终输出按行读取的字符串。

6. ZigZag Conversion

Total Accepted: 82952 Total Submissions: 349151 Difficulty: Easy

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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题目意思：简单解释一下，就是把字符串原顺序012345……按下图所示排列：

从中可以发现所有行的重复周期都是 2 * nRows - 2

对于首行和末行之间的行，还会额外重复一次，重复的这一次距离本周期起始字符的距离是 2 * nRows - 2 - 2 * i

C++实现：

class Solution {
public:
    string convert(string s, int numRows) {
        int i,j;
        int len=s.length();
        if(len==0||numRows<2)
        {
            return s;
        }
        int dis=2*numRows-2;
        string str="";
        for(i=0;i<numRows;i++)
        {
            for(j=i;j<len;j+=dis)
            {
                str+=s[j];
                if(i>0&&i<numRows-1)
                {
                    int t=j+dis-2*i;
                    if(t<len)
                    {
                        str+=s[t];
                    }
                }
            }
        }
        return str;

    }
};

python实现：

class Solution(object):
    def convert(self, s, numRows):
        """
        :type s: str
        :type numRows: int
        :rtype: str
        """
        lag=2*numRows-2
        length=len(s)
        if length==0:
            return s
        if numRows<2:
            return s
        str=""
        i=0
        while i<numRows:
              j=i
              while j<length:
                 str+=s[j]
                 if i>0 and i<numRows-1:
                    t=j+lag-2*i
                    if t<length:
                       str+=s[t]
                 j+=lag
              i+=1
        return str