Leetcode之Count and Say

最新推荐文章于 2020-04-03 01:01:57 发布

一只荣

最新推荐文章于 2020-04-03 01:01:57 发布

阅读量156

点赞数

CC 4.0 BY-SA版权

分类专栏： leetcode和机试题文章标签： leetcode

本文链接：https://blog.youkuaiyun.com/qq_35455503/article/details/86068407

leetcode和机试题专栏收录该内容

509 篇文章

订阅专栏

本文介绍了一种使用递归方法生成计数与说数序列的算法实现，该序列根据前一项生成下一项，例如从1开始，依次生成11211211111221等。文章提供了C++代码示例，展示了如何将整数和字符转换为字符串。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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 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"

代码：

class Solution {
public:
    string countAndSay(int n) {
        if (n < 1 || n>30) return "";
	if (n == 1)return "1";
	string last = countAndSay(n - 1);
	char c='.';
	string res = "";int count = 0;
	for (int i = 0; i < last.length(); i++) {
		
		if (c != last[i]) {
			if (count != 0) {
				string s1 = to_string(count);
				s1.push_back(c);
				res += s1 ;
		   }
			c = last[i];
			count = 1;
		}
		else count++;
	}

	
	string s1 = to_string(count);
	s1.push_back(c);
	res += s1 ;
	return res;
    }
};

注：

1、使用递归的思路求解

2、学会不同类型之间的转换：int 转为 string---to_string()，char转为string---s.push_back()