leetcode481.Magical String

最新推荐文章于 2025-08-13 22:03:49 发布

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最新推荐文章于 2025-08-13 22:03:49 发布

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本文介绍了一种特殊的魔法字符串，该字符串由字符‘1’和‘2’组成，并遵循特定的生成规则。通过分析其生成模式，文章提供了一种计算前N个字符中‘1’的数量的方法。

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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.

解法：

注意观察数字的规律，1和2交替出现，比如开始是122……，说明第一个数字是1出现了一次，第二个数字是2出现了两次，第三个数字是1出现了两次，就可以写出来12211，再往下写。

代码：

 class Solution {
    public:
        int magicalString(int n) {
            int count = 0;
            if(n == 0)
                return 0;
            if(n < 4)
                return 1;
            vector<int> s;
            s.push_back(1);
            s.push_back(2);
            s.push_back(2);
            int t = 2;
            while(s.size() <= n)
            {
                for(int i = 0; i < s[t]; i ++)
                {
                    if(t%2 == 0)
                        s.push_back(1);
                    else
                        s.push_back(2);
                }
                t ++;
            }
            int num = 0;
            for(int i = 0; i < n; i ++)
            {
                if(s[i] == 1)
                    num ++;
            }
            return num;
        }
    };