Leetcode 38. Count and Say Java解法

最新推荐文章于 2020-06-13 01:30:16 发布

原创最新推荐文章于 2020-06-13 01:30:16 发布 · 209 阅读

0 ·

CC 4.0 BY-SA版权

Leetcode 同时被 2 个专栏收录

32 篇文章

订阅专栏

Java

6 篇文章

订阅专栏

本文介绍了一种生成计数与描述序列的方法，通过迭代读取并描述前一项的数字特征，生成序列的第n项。示例展示了如何使用Java实现这一算法，包括递归过程和字符串操作。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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 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"

My approach

class Solution {
    public String countAndSay(int n) {
        if(n<1) return "";
        String s = "1";
        while (n>1){
            s = find(s);
            n--;
        }
        return s;
    }
    
    String find(String s){
        StringBuilder str = new StringBuilder(); //because String is immutable, concatnation cost time
        int cons = 0;
        
        for (int i = 0; i<s.length(); i++){
            if (i+1 == s.length()||s.charAt(i) != s.charAt(i+1)){
                str.append(String.valueOf(++cons));
                str.append(s.charAt(i));
                cons = 0;
            }else{ //if i<s.length()-1 &&s.charAt(i)==s.charAt(i+1)
                cons++;
            }
        }
        return str.toString(); // str is a StringBilder
    }
}

分析：

时间复杂度：O(N^2), where N is the input n.