LeetCode Top Interview Questions 38. Count and Say (Java版; Easy)

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

1.     1
2.     11   读出上一组数: 1个1, 记作11
3.     21   读出上一组数: 2个1, 记作21
4.     1211 读出上一组数: 1个2,1个1, 记作1211
5.     111221 读出上一组数: 1个1,1个2,2个1, 记作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 nth term of the count-and-say sequence.

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



Example 1:

Input: 1
Output: "1"
Example 2:

Input: 4
Output: "1211"

class Solution {
    public String countAndSay(int n) {
        //细节:这样的初始化
        StringBuilder pre;
        StringBuilder cur = new StringBuilder("1");
        int count;
        char ch;
        //自底向上
        for(int i=2; i<=n; i++){
            //用cur描述pre
            pre = cur;
            cur = new StringBuilder();
            //
            ch = pre.charAt(0);
            count = 1;
            //1ms的优化, 将for(int j=1; j<pre.length(); j++)改进成for(int j=1, len=pre.length(); j<len; j++)
            for(int j=1, len=pre.length(); j<len; j++){
                if(ch == pre.charAt(j)){
                    count++;
                }
                else{
                    cur.append(count).append(ch);
                    //update
                    ch = pre.charAt(j);
                    count = 1;
                }
            }
            //细节: 需要在内循环结束后再单独更新一次cur
            cur.append(count).append(ch);
        }
        return cur.toString();
    }
}