6. ZigZag Conversion [Z-shape converting]

Description
will be given a string based on the number given row, to down, left to right from the Z-shaped arrangement.

Example, the input string is the number of rows "LEETCODEISHIRING" is 3, arranged as follows:

LCIR
ETOESIIG
EDHN
After that, you need the output from left to right read line by line, produce a new string, such as: "LCIRETOESIIGEDHN".

Examples of

ideas

• Method 1 (I do) the string into n lines, each obtained his string, and then stitching (Because it involves a variety of computing, it is not recommended）
• Method 2

[numRows empty string] traversal s, in turn spliced to a character string of a string array (arr [index]) in, index change: from 0 to numRows, and from 0 to numRows, and so forth, Z-shaped Well
the answer

• python

*方法1*
    def convert(self, s: str, n: int) -> str:
        if n==1: return s
        ss = ''
        for r in range(1,n+1):
            if r==1 or r==n:
                j=r-1
                while j<len(s):
                    ss += s[j]
                    j+=(2*n-3+1) #n==1时不行
            else:
                j=r-1
                count=1
                while j<len(s):
                    ss += s[j]
                    if count%2==1:
                        j += 1+(n-1-r)*2+1
                    else:
                        j += 1+(r-2)*2+1
                    count+=1
        return ss
*方法2*
   def convert(self, s: str, numRows: int) -> str:
        if numRows==1:return s
        res=['']*numRows
    
        index = 0
        flag=1
        for c in s:
            res[index]+=c
            if index==numRows-1: flag=-1
            if index==0: flag=1
            index+=flag
           
      
        return ''.join(res)

• c++

*方法1*
    string convert(string s, int n) {
        if (n==1) return s;
        string ss("");
        for (int r=1; r<=n; r++)
            
        {
            if (r==1 || r==n)
            {
                int j=r-1;
                while (j<s.size())
                {
                    ss += s[j];
                    j+=2*n-3+1;
                }
            }
            else
            {
                int j=r-1,count=1;
                
                while (j<s.size())
                {
                    ss+=s[j];
                    if (count%2==1)
                        j+=1+(n-1-r)*2+1;
                    else
                        j+=1+(r-2)*2+1;
                    count++;
                }
                
            }
            
        }
        return ss;
    }
*方法2*
    string convert(string s, int numRows) {
       
        if (numRows==1) return s;
        vector<string> v(numRows, "");
        int index=0,flag=1;
        for (char c: s)
        {
            v[index].push_back(c);
            if (index==numRows-1) flag=-1;
            if (index==0) flag=1;
            index+=flag;
            /*
            if (flag==1) index++;
            if (flag==-1) index--;
            */
        }
        string res("");
        for (auto t: v)
            res+=t;
        return res;
    }