38. Count and Say

The count-and-say sequence is the sequence of integers with the first five terms as following:

1.     1
2.     11
3.     21
4.     1211
5.     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 where 1 ≤ n ≤ 30, generate the n^th term of the count-and-say sequence. You can do so recursively, in other words from the previous member read off the digits, counting the number of digits in groups of the same digit.

Note: Each term of the sequence of integers will be represented as a string.

Example 1:

Input: 1
Output: "1"
Explanation: This is the base case.

Example 2:

Input: 4
Output: "1211"
Explanation: For n = 3 the term was "21" in which we have two groups "2" and "1", "2" can be read as "12" which means frequency = 1 and value = 2, the same way "1" is read as "11", so the answer is the concatenation of "12" and "11" which is "1211".

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这题目英文描述真是醉了，看半天没理解意思，还是看网上翻译的。其实就是后一个字符串是前一个字符串的read，比如第三个字符串是21，它包含1个2和1个1，所以第四个字符串就是1211。 果然是easy模式，用递归几行代码搞定。

class Solution {
public:
    string countAndSay(int n)
    {
        if (n == 1)
            return "1";
        string pre = countAndSay(n – 1);
        string ans = "";
        int i = 0, j;
        while (i < pre.size()) {
            j = i + 1;
            while (j < pre.size() && pre[j] == pre[i])
                j++;
            ans += to_string(j – i) + pre.substr(i, 1);
            i = j;
        }
        return ans;
    }
};

本代码提交AC，用时4MS。