[LeetCode]Magical String

题目描述:

LeetCode 481. Magical String

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.

题目大意:

魔法字符串S只包含'1'和'2'并遵从以下规则:

字符串S由连续出现的'1'和'2'拼接而成,并可以由其本身生成。

S的起始部分元素为:S = "1221121221221121122……"

如果按照连续出现的'1'和'2'分组,则有:

1 22 11 2 1 22 1 22 11 2 11 22 ......

每一个组内的'1'和'2'的出现次数为:

1 2    2 1 1 2 1 2 2 1 2 2 ......

观察可见上述序列是S的一部分。

给定整数N,返回S的前N个字符中'1'的个数。

注意:N不会超过100000

解题思路:

字符串模拟

令魔法字符串ms = '122',维护指针p,初始令p = 2

若ms[p] == '1' 则向ms追加1个与ms末尾元素不同的字符

否则,向ms追加2个与ms末尾元素不同的字符

Python代码:

class Solution(object):
    def magicalString(self, n):
        """
        :type n: int
        :rtype: int
        """
        ms = '122'
        p = 2
        while len(ms) < n:
            ms += str(3 - int(ms[-1])) * int(ms[p])
            p += 1
        return ms[:n].count('1')

 

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