leetcode481. Magical String

題目要求

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.

這題是描述了一個魔法字符串，該字符串完全由數字1和2構成。這個字符串的魔法點在於，如果將該該字符串連續的數字數量進行統計並且構成一個新的字符串，會發現新的字符串與原來的字符串完全相同。
比如1 22 11 2 1 22 1 22 11 2 11 22字符串，經過統計後發現有1個1，2個2，2個1，1個2，1個1，2個2，1個1，2個2，2個1，1個2，2個1，2個2，將統計的數量合併爲新的字符串，會發現新的字符串爲1 22 11 2 1 22 1 22，和原字符串完全匹配。

思路和代碼

用雙指針可以快速的解決這個問題。假設讓一個指針指向魔法字符串中統計位，將另一個指針指向魔法字符串中的數字位，則可以不斷的通過統計位推理出當前的數字位值爲多少。如果填入數字位的值爲1，則把計數加一。代碼如下：

    public int magicalString(int n) {
        if(n == 0) return 0;
        if(n <= 3) return 1;
        int a[] = new int[n+1];
        a[0] = 1; 
        a[1] = 2; 
        a[2] = 2;
        int leftPointer = 2, rightPointer = 3, num = 1, count = 1;
        while(rightPointer < n) {
            for(int i = 0 ; i < a[leftPointer] && rightPointer < n ; i++) {
                a[rightPointer++] = num;
                if(num == 1) {
                    count++;
                }
            }
            leftPointer++;
            num ^= 3;
        }
        return count;
    }