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 text, int nRows);

convert("PAYPALISHIRING", 3) should return "PAHNAPLSIIGYIR".

思路：z字型填一个矩阵，时间复杂度为O(n^2)，注意边界值

class Solution {
public:
    string convert(string s, int numRows) {
        if(s.size()==0 || s.size()<=numRows || numRows<=1)return s;//注意边界值
        int numCols=s.size()*2/(numRows*2-2)+1;
        char **arr=(char**)malloc(sizeof(char*)*numRows);
        for(int i=0; i<numRows; ++i)//初始化矩阵
        {
            arr[i]=(char*)malloc(sizeof(char)*numCols);
            for(int j=0; j<numCols; ++j)arr[i][j]='\0';
        }
        int k=0;
        for(int j=0; j<numCols; j=j+2)//填矩阵
        {
            for(int i=0; i<numRows && k<s.size(); ++i)arr[i][j]=s[k++];
            if(j+1<numCols)
            {
                for(int i=numRows-2; i>=1 && k<s.size(); --i)arr[i][j+1]=s[k++];
            }
        }
        string str;
        for(int i=0; i<numRows; ++i)//遍历矩阵得到结果
        {
            for(int j=0; j<numCols; ++j)
            {
                if(arr[i][j]!='\0')str=str+arr[i][j];
            }
        }
        for(int i=0; i<numRows; ++i)delete []arr[i];//释放内存
        delete []arr;
        return str;
    }
};