LeetCode - Count and Say
题意
给一个字符串，他的下一个序列是每个连续的数字的个数+当前数字。求第N个序列。
思路
直接用string了。
代码
class Solution {
public:
string countAndSay(int n)
{
string str = "1";
for (int cnt = 2; cnt <= n; cnt++)
{
string curStr;
for (string::iterator it = str.begin(); it != str.end(); )
{
string::iterator ed = it;
while (ed != str.end() && *ed == *it) ed++;
int pre = count(it, ed, *it);
stringstream sstr;
sstr << pre;
curStr += sstr.str() + *it;
it = ed;
}
str = curStr;
}
return str;
}
};