[LeetCode] Zigzag Conversion

[Problem]

The string "PAYPALISHIRING" is written in a zigzagpattern on a given number of rows like this: (you may want todisplay this pattern in a fixed font for better legibility)

P   A   H   NA P L S I I GY   I   R

And then read line by line: "PAHNAPLSIIGYIR"

Write the code that will take a string and make this conversiongiven a number of rows:

string convert(string text, int nRows);

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

[Analysis]
直接看几个例子就知道该怎么做了：

ABCDEFGHIJKLMNOP 2
A C E G I K M O
B D F H J L N P

ABCDEFGHIJKLMNOP 4
A  G   M
B F H L N
C E I K O
D  J   P

ABCDEFGHIJKLMNOP 5
A   I
B H J P
C G K O
D F L N
E   M

[Solution]

class Solution {
public:
    string convert(string s, int nRows) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function   
   
        // only one row
        if(nRows == 1){
            return s;
        }
   
        // more than one rows
        int j = 0, k = 0;    // jth column, k letters has been writen
        string column;
   
        // the matrix stores the zigzag writen letters
        vector<string> matrix;
        while(k < s.length()){
            // the zigzag column, if nRows=2, it should not be zigzag writen
            if(j % 2 == 1 && nRows > 2){
                column = " ";
   
                // out of range
                if(nRows - 2 + k > s.length()){
                    column += s.substr(k, s.length() - k);
                    k += s.length() - k;
   
                    // fill remaining spaces
                    column += string(nRows - column.length(), ' ');
                }
                else{
                    column += s.substr(k, nRows - 2);
                    k += nRows - 2;
                    column += " ";
                   
                }
   
                // reverse
                reverse(column.begin(), column.end());
                matrix.push_back(column);
            }
            else{
                if(nRows + k > s.length()){
                    column = s.substr(k, s.length() - k);
                    k += s.length() - k;
                }
                else{
                    column = s.substr(k, nRows);
                    k += nRows;
                }
                matrix.push_back(column);
            }
            j++;
        }
   
        // generate result
        string res = "";
        for(int i = 0; i < nRows; ++i){
            for(j = 0; j < matrix.size(); ++j){
                // filter spaces
                if(matrix[j].size() > i && matrix[j][i] != ' '){
                    res += matrix[j][i];
                }
            }
        }
        return res;
    }
};

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