zoukankan      html  css  js  c++  java
  • leetcode 6 ZigZag Conversion

    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”.

    解决方案:
    The idea is, the first row and last row has no offset. Each element has a fixed difference of 2(nRows-1); For the rows in between, there is a incremental offset of 2;

    0 6 12 -> distance = 2(nRows-1) = 6 offset = 0
    1 5 7 11 -> offset = distance - 2 = 4
    2 4 8 10 -> offset = distance -2 -2 = 2
    3 9 -> distance = 2(nRows-1) = 6 offset = 0

    Easy to observe. There is a catch, that you need to add the offset element with previous regular element. 5 follows 1, 4 follows 2. Otherwise, you will miss the tail if there is no vertical column in the end.Looks like a CS homework:)

    class Solution {
    public:
        string convert(string s, int nRows) {
            if(s.length() == 0 || 
                s.length()/nRows < 1 ||
                nRows == 1) 
            {
                return s;
            }
            int distance = 2*(nRows-1);
            string result;
            int offset = 0;
            for (int row = 0; row < nRows; row++)
            {
                for (int index = row; index < s.length(); index += distance)
                {
                    result+=s[index];
                    if (offset != 0 && index + distance - offset < s.length())
                    {
                        result+=s[index + distance - offset];
                    }
                }
                offset += 2;
                offset = offset % distance;
            }
            return result;
        }
    };
    

    解决方案2:
    这里写图片描述
    The problem statement itself is unclear for many. Especially for 2-row case. “ABCD”, 2 –> “ACBD”. The confusion most likely is from the character placement. I would like to extend it a little bit to make ZigZag easy understood.

    The example can be written as follow:
    1.P…….A……..H…….N
    2…A..P….L..S….I…I….G
    3…..Y………I……..R

    Therefore,

    class Solution {
    public:
        string convert(string s, int numRows)
        {
    
    
        if (numRows <= 1)
            return s;
    
        const int len = (int)s.length();
        string *str = new string[numRows];
    
        int row = 0, step = 1;
        for (int i = 0; i < len; ++i)
        {
            str[row].push_back(s[i]);
    
            if (row == 0)
                step = 1;
            else if (row == numRows - 1)
                step = -1;
    
            row += step;
        }
    
        s.clear();
        for (int j = 0; j < numRows; ++j)
        {
            s.append(str[j]);
        }
    
        delete[] str;
        return s;
    }
    };

    python解决方案:
    The idea is to use the remainder (index%period) to determine which line the character at the given index will be. The period is calculated first based on nRows. A dictionary with remainder:line as key:value is then created (this can also be done with a list or a tuple). Once these are done, we simply go through s, assign each character to its new line, and then combine these lines to get the converted string.

    The code may be further shortened by using dict comprehension:

    d={i:i if i

    def convert(self, s, nRows):
        if nRows==1:
            return s
        period= 2*(nRows -1)
        lines=["" for i in range(nRows)]
        d={} # dict remainder:line
        for i in xrange(period):
            if i<nRows:
                d[i]=i
            else:
                d[i]=period-i
    
        for i in xrange(len(s)):
            lines[ d[i%period] ] +=s[i]
    
        return "".join(lines)
    
  • 相关阅读:
    如何查看RabbitMQ日志,Rabbitmq Trace日志的使用
    windows激活 RabbitMQ's Management Plugin(必须)
    UNET
    边缘检测
    Huber Loss
    深度学习之自编码器AutoEncoder(一)
    PU learning简介
    机器学习-稀疏矩阵的处理
    R语言入门-安装R和Rstuido软件
    归一化 (Normalization)、标准化 (Standardization)和中心化/零均值化 (Zero-centered)
  • 原文地址:https://www.cnblogs.com/wuyida/p/6301336.html
Copyright © 2011-2022 走看看