481 Magical String

481. Magical String

1. Question

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 stringSis the following:S= "1221121221221121122……"

If we group the consecutive '1's and '2's inS, 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 theSitself.

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

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.

2. Implementation

(1) Two Pointers

思路: 这道题的关键在于找到形成magical string的规律，这里的规律是从"122"中的第3位开始，，该位置上是数字2，先形成2个1，然后到第四位，该位置上是数字1，形成1个2，然后第5位，该位置上是1, 形成1个1，以此类推

class Solution {
    public int magicalString(int n) {
        if (n <= 0) {
            return 0;
        }

        if (n <= 3) {
            return 1;
        }

        int[] sequence = new int[n + 1];
        sequence[0] = 1;
        sequence[1] = 2;
        sequence[2] = 2;
        int head = 2, tail = 3;
        int res = 1;
        int num = 1;

        while (tail < n) {
            for (int i = 0; i < sequence[head]; i++) {
                sequence[tail] = num;
                if (num == 1 && tail < n) {
                    ++res;
                }
                ++tail;
            }
            num ^= 3;
            ++head;
        }
        return res;
    }
}

3. Time & Space Complexity

Two Pointers: 时间复杂度O(n), 空间复杂度O(n)