Leetcode_6 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 s, int numRows);

Example 1:

Input: s = “PAYPALISHIRING”, numRows = 3
Output: “PAHNAPLSIIGYIR”

Example 2:

Input: s = “PAYPALISHIRING”, numRows = 4
Output: “PINALSIGYAHRPI”
Explanation:

P I N
A L S I G
Y A H R
P I

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 s, int numRows);

Example 1:

Input: s = “PAYPALISHIRING”, numRows = 3
Output: “PAHNAPLSIIGYIR”

Example 2:

Input: s = “PAYPALISHIRING”, numRows = 4
Output: “PINALSIGYAHRPI”
Explanation:

P I N
A L S I G
Y A H R
P I

题目大致意思：就是将一个字符串竖着以Z字型排列。
思路：其实可以用一个二维vector模拟，不过是竖着模拟。注意行越界的情况，还有当字符串是空的情况。

代码：

class Solution {
public:
    string convert(string s, int numRows)
    {
        int len = s.length();
        if(numRows==1)  return s;
        if(len==0)  return "0";
        vector<vector<char> > vec;
        vec.resize(numRows);//r行
        for (int k = 0; k < numRows; ++k) {
            vec[k].resize(len);

        }
        for(int i=0;i<numRows;i++)
            for(int j=0;j<vec[0].size();j++)  vec[i][j] = '#';
        cout<<vec[0].size()<<endl;
        int row=0,col=0,cnt=0;
        while(len)
        {
            if(col%(numRows-1)==0)
            {
                vec[row][col] = s[cnt];
                row++;
            }
            else
            {
                vec[row][col] = s[cnt];
                row--;
                col++;
            }
            if(row==numRows)
            {
                row-=2;
                col++;
            }
            --len;
            ++cnt;
        }
        cout<<col<<endl;
        string ans;
        for(int i=0;i<numRows;i++)
            for(int j=0;j<=col;j++)
            {
                if(vec[i][j]!='#')  ans.push_back(vec[i][j]);
            }
        return ans;
    }
};