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.
这道题介绍了一种神奇字符串，只由1和2组成，通过计数1组和2组的个数，又能生成相同的字符串。而让我们求前n个数字中1的个数说白了其实就是让我们按规律生成这个神奇字符串，只有生成了字符串的前n个字符，才能统计出1的个数。其实这道题的难点就是在于找到规律来生成字符串，这里我们就直接说规律了，因为博主也没有自己找到，都是看了网上大神们的解法。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，参见代码如下：
解法一：
class Solution {
public:
int magicalString(int n) {
if (n <= 0) return 0;
if (n <= 3) return 1;
int res = 1, head = 2, tail = 3, num = 1;
vector<int> v{1, 2, 2};
while (tail < n) {
for (int i = 0; i < v[head]; ++i) {
v.push_back(num);
if (num == 1 && tail < n) ++res;
++tail;
}
num ^= 3;
++head;
}
return res;
}
};
下面这种解法的思路跟上面一样，但是写法上面大大的简洁了，感觉很叼！
解法二：
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');
}
};
参考资料：
https://discuss.leetcode.com/topic/74637/short-c
https://discuss.leetcode.com/topic/74917/simple-java-solution-using-one-array-and-two-pointers
LeetCode All in One 题目讲解汇总(持续更新中…)
转载请注明来源于 Grandyang 的博客 (grandyang.com)，欢迎对文章中的引用来源进行考证，欢迎指出任何有错误或不够清晰的表达。可以在下面评论区评论，也可以邮件至 grandyang@qq.com
