[Leetcode] 481. Magical String 解题报告_magicbean string 最大值-优快云博客

本文介绍了一种特殊的字符串——魔法字符串，并提出一种O(n)时间复杂度的算法来计算该字符串中前N个字符内‘1’的数量。通过扫描并记录当前重复数字的位置及数量，实现了对魔法字符串的有效生成。

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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: 6
Output: 3
Explanation: 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;
    }
};