Leetcode 481 Magical String

最新推荐文章于 2024-08-06 17:31:36 发布

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最新推荐文章于 2024-08-06 17:31:36 发布

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本文链接：https://blog.youkuaiyun.com/Neo233/article/details/85220482

本文探讨了一种特殊的魔法字符串，其特点是通过计数连续字符'1'和'2'生成自身。文章提供了一个C++代码示例，用于计算前N个字符中'1'的数量，展示了如何递增地构建此魔法字符串并计算特定字符的频率。

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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.

这个题目考察的就是简单的模拟，题意是前面的数表示的是有多少个相应的1或者是2，然后返回总共有多少个1

1）

class Solution {
public:
    int magicalString(int n) {
        string s = "122";//定义一个原始串
        for(int i = 2, k = 1; i < n ; i++,k = 3 - k){
            for(int j = 0 ; j < s[i] - '0' ;j++){//转化为数
                s += to_string(k);//以字符串的方式添加上去
            }
        }
        int res = 0;
        for(int i = 0 ; i < n ;i++){
            res += s[i] == '1';//返回其中1的个数
        }
        return res;
    }
};