LeetCode 481 Magical String
题目描述
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 Sitself.
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.
一句话题意
有一个只包含1和2的字符串，他有一个神奇的特性：
*如果将字符串按照1和2分组后，统计每组1或2出现的次数所拼接成的字符串，和这个字符串本身一样
解题思路
直接按照这个字符串的构造方法还原这个字符串s：首先初始化s = “122”,让index指向下标为2处，开始根据index指向的字符在s后面添加字符串，如果指向的是2就添加2个，如果指向的是1就添加一个，具体添加什么字符以当前s的末尾一位的字符是1还是2为准，如果s当前最后一个字符是1就添加2，反之添加1。还原好了之后用count直接计算字符串从开始到n长度处一共有多少个’1’字符。
另一种做法是，根据第三个数字2开始往后生成数字，此时生成两个1，然后根据第四个数字1，生成一个2，再根据第五个数字1，生成一个1，以此类推，生成的数字1或2可能通过异或3来交替生成，在生成的过程中同时统计1的个数即可
class Solution {
public:
int magicalString(int n) {
string s = "122";
int i = 2;
while (s.size() < n) {
s += string(s[i++] - '0', s.back() ^ 3);
}
return count(s.begin(), s.begin() + n, '1');
}
};
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