leetcode【38】Count and Say

问题描述：

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.

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".

题目大意就是数一下前一个字符串有几个对应的数字。比如1是“1”，因为1中有一个"1"，所以2是"11"，因为2中有两个1，所以3是"21"，因为3中有一个2一个1，所以4是"1211"。

解题思路：

迭代计算式每一步的结果，没有啥特别的步骤。每次遇到与之前相同的数字，就跳过，计数器tmp++。

源码：

class Solution {
public:
    string countAndSay(int n) {
        string result = "1";
        while(--n){
            string cur = "";
            for(int i=0; i<result.size(); i++){
                int tmp = 1;
                while(i+1<result.size() && result[i+1]==result[i]){
                    tmp++;
                    i++;
                }
                cur += to_string(tmp) + result[i];
                // cout<<cur<<endl;
            }
            result = cur;
        }
        return result;
    }
};