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.

这道题介绍了一种神奇字符串，只由1和2组成，通过计数1组和2组的个数，又能生成相同的字符串。而让我们求前n个数字中1的个数说白了其实就是让我们按规律生成这个神奇字符串，只有生成了字符串的前n个字符，才能统计出1的个数。其实这道题的难点就是在于找到规律来生成字符串，这里我们就直接说规律了，因为博主也没有自己找到，都是看了网上大神们的解法。根据第三个数字2开始往后生成数字，此时生成两个1，然后根据第四个数字1，生成一个2，再根据第五个数字1，生成一个1，以此类推，生成的数字1或2可能通过异或3来交替生成，在生成的过程中同时统计1的个数即可，参见代码如下：

解法一：

class Solution {
public:
    int magicalString(int n) {
        if (n <= 0) return 0;
        if (n <= 3) return 1;
        int res = 1, head = 2, tail = 3, num = 1;
        vector<int> v{1, 2, 2};
        while (tail < n) {
            for (int i = 0; i < v[head]; ++i) {
                v.push_back(num);
                if (num == 1 && tail < n) ++res;
                ++tail;
            }
            num ^= 3;
            ++head;
        }
        return res;
    }
};

下面这种解法的思路跟上面一样，但是写法上面大大的简洁了，感觉很叼！

解法二：

class Solution {
public:
    int magicalString(int n) {
        string s = "122";
        int i = 2;
        while (s.size() < n) {
            s += string(s[i++] - '0', s.back() ^ 3);
        }
        return count(s.begin(), s.begin() + n, '1');
    }
};

参考资料：

https://discuss.leetcode.com/topic/74637/short-c

https://discuss.leetcode.com/topic/74917/simple-java-solution-using-one-array-and-two-pointers