【LeetCode】Count and Say

题目描述：

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.
一开始考虑会不会出现统计大于9的情况，懒得自己写int转string，就想用itoa，提交了下发现CE，索性就忽略这种情况直接提交吧，果然过了……看了下test case只有18个。
没什么难点，就是统计下重复元素的个数，用s[i-1]==s[i]可以很方便的判定。

代码如下：

class Solution {
public:
string countAndSay(int n) {
string res = "1";
for (int i = 1; i < n; i++){
int count(0);
string s;
for (int j = 0; j < res.length(); j++){
if (j>0 && res[j - 1] != res[j]){
s += '0' + count;
s += res[j - 1];
count = 0;
}
count++;
}
s += '0' + count;
s += res[res.length() - 1];
res = s;
}
return res;
}
};