leetcode 481. Magical String 神奇字符串+有点绕的题

以你之姓@ 2022-06-03 01:42 70阅读 0赞

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.

计算题目中描述的有规律的字符串的前n个字符中‘1’的个数。先构造这个字符串（大小大于n即可），然后计算其中‘1’的个数。规律是当前i指向的数表示次数，当前字符串的末尾的数的“相反数”表示要添加的数，比如“122”，i=2指向2，字符串末尾为“2”，即在字符串后加2个1.

代码如下：

#include <iostream>
    #include <vector>
    #include <map>
    #include <set>
    #include <queue>
    #include <stack>
    #include <string>
    #include <climits>
    #include <algorithm>
    #include <sstream>
    #include <functional>
    #include <bitset>
    #include <cmath>
    
    using namespace std;
    
    class Solution
    {
    public:
        int magicalString(int n) 
        {
            string s = "122";
            int i = 2;
            while (i < n)
            {
                s += string(s[i++] - '0', s.back() == '1' ? '2': '1');
            }
    
            return count(s.begin(),s.begin()+n,'1');
        }
    };