6. ZigZag Conversion(C++)

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".

答案:

class Solution {
public:
  string convert(string s, int numRows) {
    if(numRows==1)
      return s;
    string str;
    int loop = 1 + s.size()/(2*numRows-2);
    for(int j=0;j<numRows;j++){
      for(int t=0;t<loop;t++){
        int tmp = j+2*(numRows-1)*t;
        if(j==0 || j==(numRows-1)){
          if(tmp<s.size()){
            str+=s[tmp];
          }
        }else{
          int tmp1 = tmp+2*(numRows-1-j);
          if(tmp<s.size()){
            str+=s[tmp];
          }
          if(tmp1<s.size()){
            str+=s[tmp1];
          }
        }
      }
    }
    return str;
  }
};

更好的方法:

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;
   }
};

原文地址:https://www.cnblogs.com/devin-guwz/p/6480266.html