6. ZigZag Conversion

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本文介绍了一种将字符串以Z形方式分布在多行上的算法，并实现了一个C++程序来完成这种转换。该程序能够接收输入字符串和指定的行数，然后按照Z形布局输出转换后的字符串。

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”.

int main()
{
    string s;
    int nrow;
    while (cin >> s >> nrow)
    {
        char imap[1000][1000];  //数组大小尝试出来的
        memset(imap, 0, sizeof(imap));
        int pos = 0;
        int i = 0;
        while (pos < s.size())  
        {
            if (i % 2 == 0)   //列为偶数时
            {
                for (int k = 0; k < nrow && s[pos]!='\0'; k++) //注意添加的时候，要判断s[pos]!='\0' ,如果等于的话，则也跳出了循环，因为pos此时等于s.size()了
                {
                    imap[k][i] = s[pos++];
                }
            }
            else  //列为奇数时
            {
                if(s[pos] != '\0') //判断s[pos]!='\0'
                {
                    if (nrow == 1) //nrow == 1另外考虑 ABCD 1 --> ABCD
                        imap[0][i] = s[pos++];
                    else if (nrow == 2) //nrow == 2另外考虑  ABCD 2 -->ACBD
                        for (int k = 0; k < nrow && s[pos] != '\0'; k++)
                        {
                            imap[k][i] = s[pos++];
                        }
                    else  //nrow > 2, 从nrow - 2由下往上排
                        for (int k = nrow - 2; k >= 1 && s[pos] != '\0'; k--)
                        {
                            imap[k][i] = s[pos++];
                        }
                }
            }
            i++;
        }
        s.clear();
        for (int r = 0; r < nrow; r++)
            for (int c = 0; c < i; c++)
            {
                if (imap[r][c] != 0)s.push_back(imap[r][c]);
            }
        cout << s << endl;
    }

}