[leetcode] 38. Count and Say @ python

最新推荐文章于 2024-12-16 11:20:08 发布

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最新推荐文章于 2024-12-16 11:20:08 发布

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本文链接：https://blog.youkuaiyun.com/danspace1/article/details/86571124

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原题

The count-and-say sequence is the sequence of integers with the first five terms as following:

```
1
```
```
11
```
```
21
```
```
1211
```
```
111221
```

1 is read off as “one 1” or 11.
11 is read off as “two 1s” or 21.
21 is read off as “one 2, then one 1” or 1211.

Given an integer n where 1 ≤ n ≤ 30, generate the nth term of the count-and-say sequence.

Note: Each term of the sequence of integers will be represented as a string.

Example 1:

Input: 1
Output: “1”
Example 2:

Input: 4
Output: “1211”

解法

递推法. 第n个字符串是根据第n-1个字符串读取得来的. 构造一个字典known储存已知的字符串, 然后遍历到n, 将推导的字符串存到known中, 最后返回known[n]. 定义函数读取字符, 使用双指针计算相同字符的个数, 这里的edge case是最后一个字符是单独的一个字符, 那么当curr走到最后一个index时, 读取当前字符即可.
Time: O(n*m), m是第n个字符串的长度
Space: O(1)

代码

class Solution:
    def countAndSay(self, n):
        """
        :type n: int
        :rtype: str
        """
        def readString(s):
            curr, next = 0, 1
            res = ''
            while next < len(s):
                count = 1
                while next < len(s) and s[curr] == s[next]:
                    count += 1
                    next += 1
                res += str(count) + s[curr]
                curr = next
                next += 1
            if curr == len(s) - 1:
                res += '1' + s[curr]                
            return res                
                    
            
        known = {1: '1', 2: '11'}
        if n in known:
            return known[n]
        for i in range(3, n+1):
            known[i] = readString(known[i-1])
        return known[n]