[Leetcode] 481. Magical String 解题报告
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题目:
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: 6Output: 3Explanation: The first 6 elements of magical string S is "12211" and it contains three 1's, so return 3.
思路:
不知道magic string是不是有什么规律,如果有的话,可能就可以直接用公式返回了。
我这里用的是扫描的方法:记录一个表示当前重复数字的重复数目的位置pos,以及一个当前已经重复的数字的个数num。如果发现num已经达到magic[pos]的值了,这说明我们需要换数字了,所以加入新的数字,并且将num重置为1,pos后移;否则就加入和前面相同的数字,增加num。算法的时间复杂度是O(n),空间复杂度也是O(n)。
代码:
class Solution {public: int magicalString(int n) { if (n == 0) { return 0; } vector<int> magic(1, 1); int pos = 0, num = 1, ans = 1; for (int i = 1; i < n; ++i) { if (num == magic[pos]) { // we need to swith magic.push_back(3 - magic[i - 1]); num = 1; ++pos; } else { magic.push_back(magic[i - 1]); ++num; } if (magic.back() == 1) { ++ans; } } return ans; }};
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