[LeetCode] ZigZag Conversion

ZigZag ConversionDec 6 '11

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".

 

总感觉写的很2b。

class Solution {
public:

    string convert(string s, int nRows) {
        int size = s.size();
        if (size == 0) return s;
        if (nRows == 1) return s;
        string tmp; tmp.resize(size);
        std::vector<int> ls(nRows);
        int step = 2 * nRows - 2;
        int rows = size / step;
        int left = size - rows * step;
        ls[0] = rows + (left > 0 ? 1 : 0);
        ls[nRows - 1] = rows + (left > nRows - 1 ? 1 : 0);
        for (int i = 1; i < nRows - 1; ++i) {
            ls[i] = 2 * rows + (left > i ? 1 : 0);
        }

        if (left - nRows > 0) {
            int t = left - nRows;
            int c = nRows - 2;
            while (t-- > 0) {
                ls[c--]++;
            }
        }

        std::vector<int> offset(nRows, 0);
        for (int i = 1; i < nRows; ++i) {
            offset[i] = offset[i-1] + ls[i-1];
        }
           for (int i = 0; i < rows; ++i) {
            for (int j = 0; j < nRows; ++j) {
                tmp[offset[j]++] = s[i*step + j];
            }

            for (int j = 0; j < nRows-2; ++j) {
                int l = nRows - 2 - j;
                tmp[offset[l]++] = s[i*step+nRows+j];
            }
        }
        for (int i = 0; i < min(left, nRows); ++i) {
            tmp[offset[i]++] = s[rows*step+i];
        }
        if (left > nRows) {
            for (int i = 0; i < left-nRows; ++i) {
                tmp[offset[nRows-2-i]++] = s[rows*step+nRows+i];
            }
        }
         return tmp;
    }

};

 又写了一遍～

class Solution {
public:
    string convert(string s, int nRows) {
        if (nRows == 1) return s;
        int len = s.size(); 
        if (len == 0) return string();
        
        int step = nRows + nRows - 2;
        int parts = len / step;
        int left = len - parts * step;
        vector<int> l(nRows, parts);
    	for (int i = 1; i < nRows - 1; i++) l[i]+=parts;
        if (left > nRows) {
            for (int i = 0; i < nRows; i++) l[i]++;
			left -= nRows;
			for (int i = nRows - 2; left >0 && i >= 0; i--, left--) l[i]++;
        } else {
			for (int i = 0; left >0 && i < nRows; i++, left--)
            l[i]++;
		}
        
        vector<int> idx(nRows, 0);
        idx[0] = 0;
        for (int i = 1; i < nRows ; i++)
            idx[i] = idx[i-1] + l[i-1];
        string res;
		res.resize(len);
        int r = 0;
        bool down = true;
        for (int i = 0; i < len; i++) {
            res[idx[r]++] = s[i];
            if (down) {
                if (r == nRows-1) {
                    r--;
                    down = false;
                }
                else r++;
            } else {
                if (r == 0) {
                    r++;
                    down = true;
                } else r--;
            }
        }
        return res;
    }
};