LeetCode Easy: 38. Count and Say

一、题目

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 nth term of the count-and-say sequence.

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

二、解题思路

题目的意思就是：“计数”和“说数”，后一个就是前一个的表达，如果1211，就是：1个1，1个2，两个1，写出来就是111221。给定字符串是 s，相邻的相等的子串进行计数，结果为 str（count）+str[ i ]

三、代码

#coding:utf-8
def countStr(strs):
    counts = 0
    s = ""
    i = 0

    while i in range(len(strs)):  #遍历字符串
        for j in range(i,len(strs)):    #遍历当前字符之后的字符
            if strs[j] != strs[i]:      
                counts = j - i
                s = s + str(counts) + strs[i]
                break
            if j == len(strs)-1:     # 到最后一个字符时，应该单独考虑，因为python中的range范围包左不包右
                counts = j -i + 1
                s = s + str(counts) + strs[i]
        i = i+counts
    print(s)
    return s

def countAndSay(n):
    """
    :type n: int
    :rtype: str
    """
    if n == 1:
        print(1)
        return 1
    s = '1'
    while n > 1:
        s = countStr(s)
        n -= 1
    return s

if __name__ == '__main__':
    countAndSay(10)

四、“别人”代码

http://www.cnblogs.com/chruny/p/4926290.html

class Solution(object):
    def countStr(self,s):
        count = 0;ans = "";tmp = s[0]
        for i in range(len(s)):
            if s[i] == tmp:
                count += 1
            else:
                ans += str(count) + tmp
                tmp = s[i];count = 1
        ans += str(count) + tmp
        return ans
    def countAndSay(self, n):
        """
        :type n: int
        :rtype: str
        """
        ans = '1'
        while n > 1:
            ans = self.countStr(ans)
            n -= 1
        return ans