【Leetcode】ZigZag Conversion

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".

模拟题，要特别注意nRows为1时的特殊情况处理。

class Solution {
public:
    string convert(string s, int nRows) {
        vector<string> v(nRows);
        string ret;
        bool down = true;
        int n = s.size();
        int k = 0;
        for (int i = 0; i < n; ++i) {
            if (down) {
                v[k].push_back(s[i]);
                ++k;
                if (k == nRows) {
                    k = max(0, nRows - 2);
                    if (k > 0) {
                        down = false;
                    }
                }
            } else {
                v[k].push_back(s[i]);
                --k;
                if (k == 0) {
                    down = true;
                }
            }
        }
        for (int i = 0; i < nRows; ++i) {
            int m = v[i].size();
            for (int j = 0; j < m; ++j) {
                ret.push_back(v[i][j]);
            }
        }
        return ret;
    }
};

 或者用找规律的方法，但是也需要单独考虑nRow为1的情况。

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