【leetcode】38. Count and Say（easy）

最新推荐文章于 2020-06-18 17:10:46 发布

原创最新推荐文章于 2020-06-18 17:10:46 发布 · 216 阅读

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本文介绍了一种生成计数并描述序列的算法实现，通过递归方式生成序列的每一项。文中提供了详细的C++代码示例，包括如何解析当前序列以生成下一个序列，并附带说明了字符串拼接的方法。

还是字符串题目，1 对应“1”，第二个根据上一个的“1”，为1个“1”，即“11”，第三个分析“11”，两个1，即“21”，依次分析下去，递归即可。

效率还可以吧

还是需要注意边界

字符串拼接：http://blog.youkuaiyun.com/ljp1919/article/details/48134335

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, generate the n^th term of the count-and-say sequence.

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

string countAndSayImpl(const string s)
{
	int count = 0;
	char c ;
	stringstream ss ;
	for (int i = 0; i < s.size(); ++i)
	{
		if (count == 0)
		{
			c = s[i];
			count++;
		}
		else
		{
			// c 已经有值，与s[i]比较
			if (c == s[i])
			{
				count++;
			}
			else
			{
				ss << count <<c;
				c = s[i];
				count = 1; //这个值有1个了，不能赋值为0
			}
		}
		
	}
	if (count != 0)
		ss << count << c;
	return ss.str();
}

// 38. Count and Say
string countAndSay(int n) {
	string result = "1";
	for (int i = 0; i < n - 1; ++i)
	{
		result = countAndSayImpl(result);
	}
	return result;
}