LeetCode OJ：Count and Say（数数）

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.

好奇怪的一道题目，意思是数（三）数（四）.把数字用口头语言说出来，1就是1，2前面是1就是1一个（11），3前面2个1就是（21），然后是（1211），再是（111221）以此类推。。。 当时题目看了半天没看懂，又去别的地方查了下题目是什么意思才知道了：

 class Solution {
 public:
     string countAndSay(int n) {
         string curr = "";
         if(n <= ) return curr;
         curr = "";
         for(int i = ; i < n; ++i){
             curr = convert(curr);
         }
         return curr;
     }

     string convert(const string & prev){
         //string result = "";
         stringstream result;
         char last = prev[];
         int count = ;
         int sz = prev.size();
         for(int i = ; i <= sz; ++i){//注意是<=
             if(prev[i] == last)
                 count++;
             else{
                 result << count << last;
                 // result.append(count);　这里append无法实现，因为找不到itoa，很蛋疼，只能用stream来实现
                 // result.append(last);
                 last = prev[i];
                 count = ;
             }
         }
         return result.str();
     }
 };

下面是java版本的，用了双指针，方法和上面的还是有一点不一样的：

 public class Solution {
     public String countAndSay(int n) {
         String s = String.valueOf(1);//很方便，直接就有类似itoa的api
         for(int i = 1; i < n; ++i){
             s = say(s);
         }
         return s;
     }

     public String say(String s){
         int sz = s.length();
         int p2 = 0, p1 = 0;
         int count = 0;
         String ret = new String("");
         while(p1 < sz){
             while(s.charAt(p1) == s.charAt(p2)){
                 p1++;
                 if(p1 == sz) //检查如果超过了长度就退出
                     break;
             }
             count = p1 - p2;
             ret = ret + String.valueOf(count) + s.charAt(p2);
             p2 = p1;//更新p2
         }
         return ret;
     }
 }