[LeetCode] 6. ZigZag Converesion 之字型转换字符串

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".

这道题就是看坐标的变化，找规律并分块处理。参考：爱做饭的小莹子

规律：第一行和最后一行，就是按照2n-2的顺序一点点加的。斜着那条线的字的位置是当前列j+(2n-2)-2i(i是行的index）。

Java：

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public String convert(String s, int nRows) {          if(s == null || s.length()==0 || nRows <=0)              return "";          if(nRows == 1)              return s;                       StringBuilder res = new StringBuilder();          int size = 2*nRows-2;          for(int i=0;i<nRows;i++){              for(int j=i;j<s.length();j+=size){                  res.append(s.charAt(j));                  if(i != 0 && i != nRows - 1){//except the first row and the last row                     int temp = j+size-2*i;                     if(temp<s.length())                         res.append(s.charAt(temp));                 }             }                          }          return res.toString();      }

 

Python:

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class Solution(object):     def convert(self, s, numRows):         """         :type s: str         :type numRows: int         :rtype: str         """         if numRows == 1:             return s         step, zigzag = 2 * numRows - 2, ""         for i in xrange(numRows):             for j in xrange(i, len(s), step):                 zigzag += s[j]                 if 0 < i < numRows - 1 and j + step - 2 * i < len(s):                     zigzag += s[j + step - 2 * i]         return zigzag

C++:

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class Solution { public:     string convert(string s, int nRows) {         if (nRows <= 1) return s;         string res = "";         int size = 2 * nRows - 2;         for (int i = 0; i < nRows; ++i) {             for (int j = i; j < s.size(); j += size) {                 res += s[j];                 int tmp = j + size - 2 * i;                 if (i != 0 && i != nRows - 1 && tmp < s.size()) res += s[tmp];             }         }         return res;     } };

　　

　　

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