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   NA P L S I I GY   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".

思想是：   每numRows+1个元素为一组，计算一共多少组，每一组中的最后一个元素就是中间出现的元素。画个图可以看出来，可以计算出各个元素的坐标，相当于把二维数组拉成一维的样子      ---。---。---。---。--   如上面的把字串分成5组。   注意的问题：   1.可能元素总个数少于numRows+1 或者 只分一行，只有一个元素，可直接返回class Solution {public:string convert(string s, int numRows) {if (1 == numRows || s.size() == 1)return s;int index=0;int N;//the group counts ,each group has (numRows+1) members.int strLen = s.size();string retStr(strLen + 1, '\0');int mid;//middle rowsif (numRows%2)//odd rowsmid = numRows / 2;elsemid = numRows / 2 -1;N = strLen / (numRows+1);// N may be zero when the size of the string <= numRows if (!N)return s;if (strLen % N)N += 1;for (int j = 0; j < numRows; ++j)//j is the offset in the group{for (int i = 0; i < N; ++i)//i is the offset of the group{int m = i * (numRows+1) + j;if (m < strLen){retStr[index++] = s[m];if (j == mid){int mm = i * (numRows + 1) + numRows;if (mm < strLen)retStr[index++] = s[mm];}}}}return retStr;}};