【LeetCode】38.（Java）Count and Say

题目

The count-and-say sequence is the sequence of integers with the first five terms as following:

1.   1
2.   11
3.   21
4.   1211
5.   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 where 1 ≤ n ≤ 30, generate the n th term of the count-and-say sequence.

Note: Each term of the sequence of integers will be represented as a string.

分析

给一个整型变量n，根据规律计算第n层的字符串值，编程思路：

从第1层的“1”开始计算，用两层循环实现，外层循环实现层数i的递增，内层循环实现分析计算 i-1 层。

Java实现代码：

public class countAndSay{
    public static String test(int n) {
		StringBuilder curr = new StringBuilder("1");
		StringBuilder prev;
		int count;
		char say;
		
		for(int i=1;i<n;i++) {
			prev = curr;
			curr = new StringBuilder();
			count = 1;
			say = prev.charAt(0);	//say获取prev 0 位置字符
			for(int j=1;j<prev.length();j++) {
				if(prev.charAt(j) != say) {
        //如果j位置的字符与say字符不一致，输出count和say并重置两个变量值进行新一轮计算
					curr.append(count).append(say);
					count = 1;
					say = prev.charAt(j);
				}
				else count++;//一致
			}
                        //更新curr当前层（i层）的字符串值
			curr.append(count).append(say);
		}
		return curr.toString();
	}
}