Count and Say

Problem:

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.

Analysis:

反应太慢,大概花了3天才看懂题目什么意思。觉得怎么会那么诡异。其实就是2个digit一个pattern，第一个digit代表counter，第二个代表实际的digit。有2个1那么就是21。wiki上面有公式，但是看不懂。下面用英文清理一下思路，能用这个思路写代码。
1.Check the previous strings' consecutive digit, keep the count.
2.If the current digit is different than the previous digit

•    Push the counter and digit to the result string. 
•    Reset counter.

Solution:
（抄得别人代码）

1:  private static String countandSay(int n)  
2:       {  
3:            if (n<=0) return null;   
4:              StringBuilder res = new StringBuilder("1");   
5:              while (n-- > 1) {   
6:               StringBuilder temp = new StringBuilder();   
7:               int count = 1;   
8:               for (int i=1; i<res.length(); ++i) {   
9:                if (res.charAt(i) == res.charAt(i-1)) {   
10:                 count++;   
11:                } else {   
12:                 temp.append(count);   
13:                 temp.append(res.charAt(i-1));   
14:                 count = 1; // reset   
15:                }   
16:               }   
17:               temp.append(count);   
18:               temp.append(res.charAt(res.length()-1));   
19:               res = temp;   
20:              }   
21:              return res.toString();   
22:       }