Leetcode 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 nth 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.

1 迭代

先写一个read函数, 返回一个字符串的读音

然后调用read函数n - 1次, 最后返回结果. (但不知道为什么这么慢. faster than 5.04% of C++ online submissions for Count and Say.)

可能的改进点有, 把to_string函数换成强制类型转换, 先计算k = text.size()然后后面全部用k代替size()

class Solution {
public:
    string countAndSay(int n) {
        string ans = "1";
        while(--n)
        {
            ans = read(ans);
        }
        return ans;
    }
private:
    string read(string text)
    {
        string ans = "";
        int begin = 0;
        int end = 0;
        while(end < text.size())
        {
            while(end < text.size() && text[end] == text[begin])
                ++end;
            ans = ans + to_string(end - begin) + text[begin];
            begin = end;
        }
        return ans;
    }
};