leetcode-easy-string- 38 Count and Say

Posted 2022-01-02 rosyyy

mycode   91.28%

思路：题意实在太难理解了，尤其是英文又不好，只能参看下别人的资料，理解下规则。终于理解，题意是n=1时输出字符串1；n=2时，数上次字符串中的数值个数，因为上次字符串有1个1，所以输出11；n=3时，由于上次字符是11，有2个1，所以输出21；n=4时，由于上次字符串是21，有1个2和1个1，所以输出1211。依次类推，写个countAndSay(n)函数返回字符串。

class Solution(object):
    def countAndSay(self, n):
        """
        :type n: int
        :rtype: str
        """
        from collections import Counter
        if n == 1:
            return ‘1‘     
        a = self.countAndSay(n-1)
         
        res , temp , count = ‘‘ , a[0] , 1
        for i in range(1,len(a)):
            if a[i] == temp:
                count += 1
            else:
                res += str(count) + str(temp)
                temp = a[i]
                count = 1
        res += str(count) + str(temp)
        return res
       

 

参考

把递归换成了for循环

class Solution(object):
    def countAndSay(self, n):
        """
        :type n: int
        :rtype: str
        """        
        x= ‘1‘
        y=‘‘
        prev = ‘1‘
        for i in range(n-1):
            count = 0
            y = ‘‘
            prev = x[0]
            for j in x:

                if prev==j:
                    count+=1

                else:
                    y+=(str(count)+str(prev))
                    prev = j
                    count = 1


            y+= (str(count)+str(prev))
            print(‘y: ‘,y)
            x = y


        return(x)