481. Magical String

最新推荐文章于 2020-04-27 11:02:21 发布

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最新推荐文章于 2020-04-27 11:02:21 发布

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本文介绍了一种特殊的魔法字符串，该字符串由'1'和'2'组成，并遵循特定的生成规则。文章提供了一个算法实例，用于计算给定长度的魔法字符串中'1'的数量。

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A magical string S consists of only '1' and '2' and obeys the following rules:

The string S is magical because concatenating the number of contiguous occurrences of characters '1' and '2' generates the string S itself.

The first few elements of string S is the following:S = "1221121221221121122……"

If we group the consecutive '1's and '2's in S, it will be:

1 22 11 2 1 22 1 22 11 2 11 22 ......

and the occurrences of '1's or '2's in each group are:

1 2 2 1 1 2 1 2 2 1 2 2 ......

You can see that the occurrence sequence above is the S itself.

Given an integer N as input, return the number of '1's in the first N number in the magical string S.

Note:N will not exceed 100,000.

Example 1:

Input: 6
Output: 3
Explanation: The first 6 elements of magical string S is "12211" and it contains three 1's, so return 3.

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计算题目中描述的有规律的字符串的前n个字符中‘1’的个数。先构造这个字符串（大小大于n即可），然后计算其中‘1’的个数。规律是当前i指向的数表示次数，当前字符串的末尾的数的“相反数”表示要添加的数，比如“122”，i=2指向2，字符串末尾为“2”，即在字符串后加2个1.

代码：

class Solution
{
public:
	int magicalString(int n)
	{
		string s = "122";
		int i = 2;
		while(s.size() < n)
		{
			s += string(s[i++] - '0', s.back() == '1' ? '2' : '1');
		}
		return count(s.begin(), s.begin() + n, '1');
	}
};