problem：
The count-and-say sequence is the sequence of integers beginning as follows:
1, 11, 21, 1211, 111221, ...
1 is read off as "one
1" or 11.
11 is read off as "two
1s" or 21.
21 is read off as "one
2, then one 1" or 1211.
Given an integer n, generate the nth sequence.
Note: The sequence of integers will be represented as a string.
String
thinking：
（1）首先要读懂计数 - 读数的规律，比较有意思
（2）对重复的数计数，单个的数读一个什么数。
code：
class Solution {
public:
string countAndSay(int n) {
// string start = int_to_string(n);
string start ="1";
for(int i=0;i<n-1;i++ )
start = factory(start);
return start;
}
protected:
string factory(string str)
{
int count=1;//初始化为1 ，很巧妙
string res;
if(str.empty())
return str;
if(str.size()==1)
{
res.push_back('1');
res+=str;
return res;
}
for(string::size_type i=0;i<str.size()-1;i++)//计数到倒数第二个
{
if(str.at(i)==str.at(i+1))
{
count++;
if(i==str.size()-2) //到了末尾，别忘了存储
{
char a=count+'0';
res.push_back(a);
res.push_back(str.at(i));
}
}
else
{
char a=count+'0';
res.push_back(a);
res.push_back(str.at(i));
count=1;
}
}//size()-1
if(str.at(str.size()-1)!=str.at(str.size()-2))//处理最后一个字符
{
res.push_back('1');
res.push_back(str.at(str.size()-1));
}
// factory(res,n--);
return res;
}
/*
string int_to_string(int n) // int invert into string
{
vector<char> array;
string res;
if(n<0)
n=-n;
while(n>0)
{
int a = n%10;
n/=10;
char b = '0'+a;
array.push_back(b);
}
for(int i=array.size()-1;i>=0;i--)
res.push_back(array.at(i));
return res;
}
*/
};
leetcode 题解 || Count and Say 问题
原文：http://blog.csdn.net/hustyangju/article/details/44624139