leetcode 38. Count and Say

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, generate the n^th 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"

class Solution(object):
    def countAndSay(self, n):
        """
        :type n: int
        :rtype: str
        """
        """
        1=>1
        """
        k = "1"
        
        def say(m):
            # greedy
            ans = ""
            c = 1
            for i in xrange(1, len(m)):
                if m[i] == m[i-1]:
                    c += 1
                else:
                    ans += str(c) + m[i-1]
                    c = 1
            ans += str(c) + m[-1]
            return ans
            
        for i in xrange(1, n):
            k = say(k)
        return k
        

 经典的字符计数问题！或者也可以使用贪心算法求解，还可以避免最后为len-1的判断。

class Solution(object):
    def countAndSay(self, n):
        """
        :type n: int
        :rtype: str
        """
        """
        1=>1
        """
        k = "1"
        
        def say(m):
            # greedy
            ans = ""     
            i = 0
            while i<len(m):
                s, c = m[i], 1
                while i+1<len(m) and m[i+1] == m[i]:
                    i += 1
                    c += 1
                ans += str(c)+str(s)
                i += 1
            return ans
            
        for i in xrange(1, n):
            k = say(k)
        return k