Count and Say @LeetCode

题目比较不直观，这里的描述比较好一些 http://www.careercup.com/question?id=4425679

"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<s.length(); i++){
				if(s.charAt(i) == s.charAt(i-1)){		// 重复的值，继续计数
					cnt++;
				}else{			// 有新的值出现，记录到ret
					ret.append(cnt).append(s.charAt(i-1));
					cnt = 1;		// 重置cnt
				}
			}
			ret.append(cnt).append(s.charAt(i-1));
			s = ret.toString();	// 更新s
		}
		return ret.toString();
	}

}