38.leetcode38 Count and Say

本题难度为easy,给了5个字符串，根据规律推测出后续行数的字符，题目要求输入行数返回当行字符。答案里面讲输入行数分成了3类，

1.如果为0返回空字符

2.如果小于等于5直接返回已有字符

3.大于5后开始每行按照规律计算，直到输入行数并返回值

38. Count and Say

Easy

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

1.     1
2.     11
3.     21
4.     1211
5.     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. You can do so recursively, in other words from the previous member read off the digits, counting the number of digits in groups of the same digit.

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

 

Example 1:

Input: 1
Output: "1"
Explanation: This is the base case.

Example 2:

Input: 4
Output: "1211"
Explanation: For n = 3 the term was "21" in which we have two groups "2" and "1", "2" can be read as "12" which means frequency = 1 and value = 2, the same way "1" is read as "11", so the answer is the concatenation of "12" and "11" which is "1211".

 

 

class Solution(object):
    def countAndSay(self, n):
        """
        :type n: int
        :rtype: str
        """
        origin_list = [[1],[1,1],[2,1],[1,2,1,1],[1,1,1,2,2,1]]
        if n == 0 :
            return ""
        if n<=5:
            rtn = ""
            for item in origin_list[n-1]:
                rtn+=str(item)
            return rtn
                
        else:
            for i in range(5,n):
                last_row = origin_list[i-1]
                this_row = []
                count = 0
                cur_num = last_row[0]
                for data in last_row:
                    if data!=cur_num:
                        this_row.extend([count,cur_num])
                        count = 1
                        cur_num = data
                    else:
                        count+=1
                this_row.extend([count,cur_num])
                origin_list.append(this_row)
            rtn = ""
            for item in origin_list[-1]:
                rtn+=str(item)
            return rtn