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.

// 理解了題意之後，發現這樣的magical string是需要構造的，因此，首先根據它的性質把這樣的字串構造出來，再進行統計。這樣比較清晰直接
// 具體構造過程見程式碼
class Solution {
public:
    int magicalString(int
 
 n) {
        if (n == 0) return 0;
        // 只有1和2，用vector表示比string方便。
        //構造時一次性可能增加兩個元素，索性就把vector稍微開大點
        vector<int> res(n + 5, 0); 
        // 先給定三個初值，再進行構造比較方便
        //（原本想只給定一個初值就進行構造，但程式碼寫起來總覺得怪怪的，索性直接給三個吧）
        res[0] = 1;
        res[1] = 2;
        res[2] = 2;
        // 兩個指標，ir指向準備當前已構造好的末尾位置，ig指向的數字幫助我們進行構造，ig<=ir
        int ir = 2, ig = 2;
        while (ir < n) {
            // 從magical string的性質發現，每次構造要麼能直接構造兩個重複的數，要麼只能構造一個
            // 且新構造的數總是和前面的數不同
            res[ir+1] = (res[ir] == 1) ? 2 : 1;
            if (res[ig] == 2) {
                res[ir+2] = res[ir+1];
                ir += 2;
            } else {
                ir += 1;
            }
            ++ig;
        }
        // 最後進行1個數的統計，這裡用了for_each和count_if，還有lambda表示式
        // 最近正好在補充相關知識，拿出來練習一下
        int cnt = 0;
        //for_each(res.begin(), res.begin()+n, [&cnt](int x) {if (x == 1) ++cnt;});
        cnt = count_if(res.begin(), res.begin()+n, [](int x) {return (x == 1);});
        return cnt;
    }
};