leetcode：Count and Say

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.

题目意思是n=1时输出字符串1；n=2时，数上次字符串中的数值个数，因为上次字符串有1个1，所以输出11；n=3时，由于上次字符是11，有2个1，所以输出21；n=4时，由于上次字符串是21，有1个2和1个1，所以输出1211。依次类推（不断数数），写个countAndSay(n)函数返回字符串。

class Solution {
public:
    string convert(const string &say)
    {
       stringstream ss;    //创建一个流
       int count=0;
       char last=say[0];
       for(size_t i=0; i<=say.size(); ++i)
       {
           if(last==say[i])  //每次与前一个字符比较是否相等，相等则计数加一
           {
               ++count;
           }
           else
           {
           ss<<count<<last;   //将count的值传递到流ss中，再将last的值传入
           count=1;
           last=say[i];
           }
       }
       return ss.str();   //返回流中的字符串
    }
    string countAndSay(int n) {
        if(n<=0) return string();
        string say="1";    //定义初始字符串的内容
        for (int i=1; i<n; ++i)
        {
            say=convert(say);
        }
        return say;
    }
};

　或：

class Solution {
public:
string nextRead(string s) {
    stringstream ss;
    int count, i = 0, n = s.length();
    while (i < n) {
        count = 0;
        while (i + 1 < n && s[i] == s[i + 1]) {
            i++;
            count++;
        }
        ss << count + 1 << s[i];
        i++;
    }
    return ss.str();
}
string countAndSay(int n) {
    string res = "";
    if (n == 0) return res;
    res = "1";
    if (n == 1) return "1";
    while (n > 1) {
        res = nextRead(res);
        n--;
    }
    return res;
}
};

　　其他解法：

class Solution {
public:
   string countAndSay(int n) {
    if (n==0){
        return NULL;
    }
    if (n==1){
        return "1";
    }
    int count=1;
    string s1="1";
    string s2;
    int j=1;
    if (n>=2){
        s1="11";
        j=2;
    }
    int i=0;
    while (j<n && n>1){
        s2="";
        for (i=0;i<s1.size();i++){

            for (int m=i;m<s1.size()-1;m++){
                if (s1[m]==s1[m+1]){
                  count++;
                  i++;
                }
                if (s1[m]!=s1[m+1]){
                break;
                 }
            }
                s2+=to_string(count)+s1[i];
                count=1;

        }
        s1=s2;
        j++;
    }

    return s1;
}
};

　　

class Solution {
public:
 string countAndSay(int n) {
    if(n==0)  return " ";
    if(n==1)  return "1";
    string str="1";
    string tmp;
    for(int i=1;i<n;++i)
    {
        tmp.clear();
        int cnt=1;
        for(int j=0;j<str.size();++j)
        {
            while(j+1<str.size())
            {
                if(str[j]==str[j+1])
                {
                    ++cnt;
                    ++j;
                }
                else
                break;
            }
            tmp.push_back(cnt+'0');
            tmp.push_back(str[j]);
            cnt=1;
        }
        str=tmp;
    }
    return str;
}
};