題目名稱：6. ZigZag Conversion

題目難度：Medium

題目描述：

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

string convert(string text, int nRows);
convert(PAYPALISHIRING, 3) should return PAHNAPLSIIGYIR.

題目分析：

題目並沒有說清楚zigzag樣式到底是一個什麼樣的樣式，查了一下才知道是這樣子的：

Δ=2n-2    1                           2n-1                         4n-3
Δ=        2                     2n-2  2n                    4n-4   4n-2
Δ=        3               2n-3        2n+1              4n-5       .
Δ=        .           .               .               .            .
Δ=        .       n+2                 .           3n               .
Δ=        n-1 n+1                     3n-3    3n-1                 5n-5
Δ=2n-2    n                           3n-2                         5n-4
 

於是就有了思路，很直接地用一個二維矩陣儲存起來，然後再按要求輸出就行，比較需要注意的是矩陣的大小，可以使用壓縮處理將中間一些比較稀疏的列合在一起。
最後AC的程式碼是：

class Solution {
public:
    string convert(string s, int numRows) {
        if (s.size() <= 1 || numRows == 1) return s;
        int numCols;
        if (s.size() <= numRows) {
            numCols = 1;
        } else
 
 {
            int t = numRows * 2 - 2;
            int temp = s.size() / t * 2;
            int toIf = s.size() % t;
            if (toIf < numRows) numCols = temp + 1;
            else numCols = temp + 2;
        }
        vector<char> temp(numCols, '-');
        vector<vector<char>> v;
        v.assign(numRows, temp);
        int index = 0;
        bool flag = 0;
        for (int j = 0; j < numCols; ++j) {
            if (j % 2 == 0) {
                for (int i = 0; i < numRows; ++i) {
                    v[i][j] = s[index++];
                    if (index >= s.size()) {
                        flag = 1;
                        break;
                    }
                }
            } else {
                for (int i = numRows - 2; i > 0; --i) {
                    v[i][j] = s[index++];
                    if (index >= s.size()) {
                        flag = 1;
                        break;
                    }
                }
            }
            if (flag) break;
        }
        string result;
        for (int i = 0; i < numRows; ++i) {
            for (int j = 0; j < numCols; ++j) {
                if (v[i][j] != '-') {
                    result += v[i][j];
                }
            }
        }
        for (int i = 0; i < numRows; ++i) {
            for (int j = 0; j < numCols; ++j) {
                cout << v[i][j] << ' ';
            }
            cout << endl;
        }
        cout << endl;
        return result;
    }
};