LeetCode - ZigZag Conversion

LeetCode - ZigZag Conversion

2013.12.1 21:48

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:

1 string convert(string text, int nRows);

convert("PAYPALISHIRING", 3) should return "PAHNAPLSIIGYIR".

Solution:

　　This kind of problem is not complicated, and can be done with some calculation on the scratch before you start coding. Still, special cases need special handlings:

　　　　1. the string is empty

　　　　2. nRows = 1

　　Time complexity is O(n), where n is the length of the string. Space complexity is O(n) with the use of char array during the algorithm.

Accepted code:

 1 //1RE, 1AC, special case needs special handling

 2 class Solution {

 3 public:

 4     string convert(string s, int nRows) {

 5         // IMPORTANT: Please reset any member data you declared, as

 6         // the same Solution instance will be reused for each test case.    

 7         

 8         // 1RE here, ignored the special case where nRows == 1

 9         if(nRows == 1){

10             return s;

11         }

12         char *str = nullptr;

13         int i, j, k, len;

14         

15         len = s.length();

16         str = new char[s.length() + 1];

17         j = 0;

18         for(k = 0; k < nRows; ++k){

19             if(k == 0){

20                 for(i = 0; i < len; ++i){

21                     if(i % (2 * nRows - 2) == 0){

22                         str[j++] = s[i];

23                     }

24                 }

25             }else if(k == nRows - 1){

26                 for(i = 0; i < len; ++i){

27                     if(i % (2 * nRows - 2) == nRows - 1){

28                         str[j++] = s[i];

29                     }

30                 }

31             }else{

32                 for(i = 0; i < len; ++i){

33                     if(i % (2 * nRows - 2) == k || i % (2 * nRows - 2) == 2 * nRows - 2 - k){

34                         str[j++] = s[i];

35                     }

36                 }

37             }

38         }

39         str[j] = 0;

40         

41         string res = string(str);

42         delete[] str;

43         return res;

44     }

45 };