leetcode38:Count and Say

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思路：对每个字符串判断连续相同为的个数，然后进行拼接即可。

代码：

public class CountAndSay {

	public static void main(String[] args) {
		System.out.println(countAndSay(1));
	}

	public static String countAndSay(int n) {

		String s = "1";
		StringBuilder sb = null;
		int count = 1;
		for (int i = 2; i <= n; i++) {
			sb=new StringBuilder();
			count=1;
			for (int j = 0; j < s.length(); j++) {
				if (j < s.length() - 1)
					if (s.charAt(j) == s.charAt(j + 1))
						count++;
					else {
						sb = sb.append(count).append(s.charAt(j));
						count = 1;
					}
				else
					sb=sb.append(count).append(s.charAt(j));
			}
			s=sb.toString();
			//System.out.println(s);
		}
		return s;

	}
}

输出：