ZigZag Conversion--LeetCode

1.题目

ZigZag Conversion

2.代码

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