Count and Say @LeetCode

"Count and Say problem" Write a code to do following:

n String to print

0 1

1 1 1

2 2 1

3 1 2 1 1

...

Base case: n = 0 print "1"

for n = 1, look at previous string and write number of times a digit is seen and the digit itself. In this case, digit 1 is seen 1 time in a row... so print "1 1"

for n = 2, digit 1 is seen two times in a row, so print "2 1"

for n = 3, digit 2 is seen 1 time and then digit 1 is seen 1 so print "1 2 1 1"

for n = 4 you will print "1 1 1 2 2 1"

Consider the numbers as integers for simplicity. e.g. if previous string is "10 1" then the next will be "1 10 1 1" and the next one will be "1 1 1 10 2 1"

package Level2;  
  
/** 
 * 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. 
 * 
 */  
public class S38 {  
  
    public static void main(String[] args) {  
        System.out.println(countAndSay(6));  
    }  
      
    public static String countAndSay(int n) {  
          
        if(n == 1){  
            return "1";  
        }  
          
        String s = "1";  
        StringBuffer ret = new StringBuffer();  
        int cnt = 0;  
        int round = 0;          // round是迭代多少次  
        int i;  
        while(++round < n){  
            cnt = 1;  
            ret.setLength(0);  
            for(i=1; i