第七周算法分析与设计：ZigZag Conversion

算法描述：

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"
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".

题目来自此处

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解决方案：
我的解决思路是（贼麻烦）：二维char数组，以'Z'型的方式存储字符(一开始字符数组以'0'填充)，然后最后通过检测不是'0'的字符存储进要输出字符串中。

string convert(string s, int numRows) {    if (numRows == 1) return s;    if (s == "") return "";    char **ch = new char*[numRows];    int i, j, k = 0;    for (i = 0; i<numRows; i++){        ch[i] = new char[s.length() + 1];    }    for (i = 0; i<numRows; i++){        for (j = 0; j <= s.length(); j++){                ch[i][j] = '0';             }    }    int count = s.length();    i = 0;    j = 0;    while (k<count){        if (i == 0){            if (j != 0) i++;            while (i<numRows && k<count){                ch[i][j] = s[k];                i++; k++;            }        }        else if (i >= numRows){            ch[numRows - 2][j + 1] = s[k];            i = numRows - 2;            j++;            k++;        }        else{            i--;            j++;            ch[i][j] = s[k];            k++;        }    }    string rs = "";    for (i = 0; i<numRows; i++){        for (j = 0; j <= s.length(); j++){            if(ch[i][j]!='0'){                  rs += ch[i][j];             }        }    }    return rs;    }

这么麻烦的写法时间和空间复杂度都是O(n2)……学习答案思路去。。