#38. Count and Say

Work_Life

于 2017-09-06 13:03:58 发布

阅读量218

点赞数

CC 4.0 BY-SA版权

分类专栏： Leetcode刷题

本文链接：https://blog.youkuaiyun.com/Work_Life/article/details/77865117

Leetcode刷题专栏收录该内容

13 篇文章

订阅专栏

本文介绍了一种基于逻辑模拟的方法来生成计数猜数序列的第n项。通过递归迭代的方式，从初始值1开始，逐步构建出完整的序列。每一步都详细记录了当前字符串的构成方式及其读法。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

题目描述：

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

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 term of the count-and-say sequence.

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

找规律模拟的题，考逻辑。

class Solution {
public:
    string countAndSay(int n) {
        if(n==1)
            return "1";
        else
            return iteration("1",1,n);
    }
    string iteration(string s,int k,int n){
        //cout<<"k = "<<k<<" "<<"s = "<<s<<endl;
        if(k==n)
            return s;
        string result;
        int i,j;
        for(i=0;i<s.size();){
            j=i;
            while(j<s.size()&&s[i]==s[j])
                j++;
            result.push_back(j-i+'0');
            result.push_back(s[i]);
            i=j;
            //cout<<"result = "<<result<<endl;
        }
        return  iteration(result,k+1,n);
    }
};

本身是一个迭代的过程