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".

class Solution {
public:
    string convert(string s, int nRows) {
        int len = s.length();
        if (len < 2 || nRows == 1 || len < nRows) return s;
        vector<string> cols;

        int si = 0;
        int ri = 0;
        
        while (si < len) {
            bool inacol = cols.size() % (nRows-1) == 0;
            
            if (!inacol) {
                for (int i=0; i<nRows - 2 && si <len; i++, si++) {
                    cols.push_back(s.substr(si, 1));      // a char as a single column
                }
                continue;
            }
            
            cols.push_back(string());

            for (int i=0; i<nRows && si<len; i++, si++) {
                cols.back().push_back(s[si]);   // all char in a column
            }
        }

        string res;
        int nCols = cols.size();

        int stepa = nRows - 1;
        int stepb = 0;
        
        for (int i=0; i<nRows; i++) {
            bool usea = false;
            int last = -1;
            for (int j = 0; j < nCols; j += usea ? stepa : stepb) {
                usea = !usea;
                if (j == last) continue;
                last = j;
                
                int r = i, c = j;
                
                if (cols[c].length() < 1) break;
                
                if (c % (nRows - 1) != 0) {
                    r = 0;
                } else if (cols[c].length() - 1 < r) {
                    break;
                }
                res.push_back(cols[c][r]);
            }
            --stepa, ++stepb;
        }
        cols.clear();
        return res;
    }
};

一般来说那么长，时间又是100ms+的代码定不是好的！

发现一个以前没注意到的问题，如下代码：

    string a("abc");
    string b;
    b.push_back(a[10]);
    cout<<b.length()<<endl;

编译运行并不会报错，虽然其中a[10]的操作明显越界了，但是与数组不同，字符串的下标操作当越界时会返回一个NULL字符，去http://www.cplusplus.com/查了一下，上面是这么说的：

If pos is equal to the string length, the function returns a reference to a null character (charT()).

从实际测试来看，不仅仅是当str[pos]中的pos == str.length()时, 当其大于字符串长度时也是如此。虽然这样的过程是一番好意，但是有时候却让错误变得更难以发现，如下代码：

    string a("abc");
    string b(a);
    b.push_back(a[10]);
    
    cout<<a<<","<<b<<endl;
    cout<<(a == b)<<endl;

输出a, b时，他们是完全一样的，然而因为b比a多了NULL，在等值比较时就不相等了，非常令人困惑。当然根源还是，码农不小心。。。

第二轮：

居然写出如此冗长的代码，来个简短版的：

class Solution {
public:
    string convert(string s, int numRows) {
        string res;
        if (numRows < 1) {
            return res;
        }
        vector<string> z(numRows);
        int len = s.size();
        int row = 0;
        int delta = numRows == 1 ? 0 : 1;
        for (int i=0; i<len; i++) {
            z[row].push_back(s[i]);
            int next = row + delta;
            if (next == numRows || next == -1) {
                delta = -delta;
            }
            
            row += delta;
        }
        
        for (string& s : z) {
            res.append(s);
        }
        return res;
    }
};

 再来一个不用额外空间的

class Solution {
public:
    string convert(string s, int numRows) {
        string res;
        if (numRows < 1) {
            return res;
        }
        int len = s.size();
        if (len <= numRows || numRows == 1) {
            return s;
        }
        int delta = 2 * numRows - 2;
        
        for (int i=0; i<numRows; i++) {
            res +=s[i];
            int next = i + delta;
            while (next - 2 * i < len) {
                if (i != 0 && i != numRows - 1) {
                    res += s[next - 2 * i];
                }
                if(next < len) {
                    res += s[next];
                }
                next += delta;
            }
        }
        return res;
    }
};

快很多