LeetCode

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:

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:

//1RE, 1AC, special case needs special handling
class Solution {
public:
    string convert(string s, int nRows) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.    
        
        // 1RE here, ignored the special case where nRows == 1
        if(nRows == 1){
            return s;
        }
        char *str = nullptr;
        int i, j, k, len;
        
        len = s.length();
        str = new char[s.length() + 1];
        j = 0;
        for(k = 0; k < nRows; ++k){
            if(k == 0){
                for(i = 0; i < len; ++i){
                    if(i % (2 * nRows - 2) == 0){
                        str[j++] = s[i];
                    }
                }
            }else if(k == nRows - 1){
                for(i = 0; i < len; ++i){
                    if(i % (2 * nRows - 2) == nRows - 1){
                        str[j++] = s[i];
                    }
                }
            }else{
                for(i = 0; i < len; ++i){
                    if(i % (2 * nRows - 2) == k || i % (2 * nRows - 2) == 2 * nRows - 2 - k){
                        str[j++] = s[i];
                    }
                }
            }
        }
        str[j] = 0;
        
        string res = string(str);
        delete[] str;
        return res;
    }
};