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:

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

思路

根据题意，利用一个初始的种子数组，生成所需要的长度为n的magic数组。
然后再利用简单的统计技巧，将所有范围内的元素加起来。其和与元素个数的差值就是2的个数；1的个数也很容易得出。

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class Solution(object):
    def magicalString(self, n):
        """
        :type n: int
        :rtype: int
        """
        seed = [1,2,2]
        cur = 2
        while( len(seed) < n ):
            if (seed[cur] == 2):
                if (seed[-1] == 2):
                    seed.append(1)
                    seed.append(1)
                elif (seed[-1] == 1):
                    seed.append(2)
                    seed.append(2)
            elif (seed[cur] == 1):
                if (seed[-1] == 2):
                    seed.append(1)
                elif (seed[-1] == 1):
                    seed.append(2)
            cur += 1
        
        ans = 0
        for i in seed[:n]:
            ans += i
        return (2 * n - ans)