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  • [LeetCode] Magical String 神奇字符串

    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...)

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  • 原文地址:https://www.cnblogs.com/grandyang/p/6286540.html
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