[leetcode]Count and Say

最新推荐文章于 2019-05-10 00:17:00 发布

原创最新推荐文章于 2019-05-10 00:17:00 发布 · 406 阅读

0 ·

CC 4.0 BY-SA版权

leetcode 专栏收录该内容

9 篇文章

订阅专栏

本文介绍如何生成给定整数n时的说计数序列，该序列通过读取前一项并描述其组成来形成。实现了一个函数，接受整数n作为输入，并返回第n项的字符串表示。

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 beginning as follows:
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 sequence.

Note: The sequence of integers will be represented as a string.

此题没找到简单的方法

n-1时候生成的former 字符串，一次读出其中的每个字符，比较former[i]与former[i+1]是否相同，不相同依次写入 count 和数即可

string countAndSay(int n) {
    string former="1";
    for(int i=1; i<n; i++){
        unsigned long length=former.length();
        int k=0;
        string temp;
        int count=1;
        while(k<length){
            if(former[k]!=former[k+1]){
                temp+='0'+count;
                temp+=former[k];
                count=1;
            }
            else{
                count++;
            }
            k++;
        }
        former=temp;
    }
    return former;
}