class Solution {
public:
string convert(string s, int numRows) {
string result = “”;
if(numRows == 1 || s.size() <= numRows) return s;
for(int i = 0;i < numRows;i++)
{
int start = i;
int step;
if((i == numRows-1) or (i == 0))
{
step = 2 * (numRows - 1);
while(start < s.size())
{
result += s[start];
start += step;
}
}
else
{
step = 2 * (numRows -1 -i);
while(start < s.size())
{
result += s[start];
start += step;
step = 2* numRows -step-2;
}
}
}
return result;
}
};
行首位每次寻找该行下一个输出字符,步长一直是2*(numRows - 1)