写一个函数,输入一个字符串,得到满足规则的下一个字符串。重复这个函数n次就可以了。 代码(python): View Code
classSolution(object):defcountAndSay(self,n):""" :type n: int :rtype: str """ifn<0:return""elifn==1:return"1"else:s=self.countAndSay(n-1)result=[]tmp=s[0]count=0foriinrange(len(s)):ifs[i]==tmp:count+=1else:result.append(str(count))result.append(tmp)count=1tmp=s[i]...
count and say 解法, 采用动态规划 class Solution(object): def countAndSay(self, n): """ :type n: int :rtype: str """ dp = [] dp.append('1') for i in range(1, n): dp.append(self.helper(dp[-1])) return dp[-1] def helper(self, s): s_r = '' j = 0 for i in ...
AC代码(Python) 1_author_ ="YE"2#-*- coding:utf-8 -*-34classSolution(object):5defcountAndSay(self, n):6"""7:type n: int8:rtype: str9"""10L = ['1']11ifn == 1:12return'1'13else:14foriinrange(n - 1):15j =016whilej <len(L):17x =L[j]18count = 119ifj + count...
class Solution: # @return a string def count(self,s): t=''; count=0; curr='#' for i in s: if i!=curr: if curr!='#': t+=str(count)+curr curr=i count=1 else: count+=1 t+=str(count)+curr return t def countAndSay(self, n): s='1' for i in range(2,n+1): s=...
Problem link: Explore - LeetCodeleetcode.com/explore/interview/card/top-interview-questions-easy/127/strings/886/ Solutions: Compare to the previous number, if the current number is the same as the previous one, then count += 1, else reset count variable and push the count and previous...
Python实现: 代码语言:javascript 复制 classSolution:defcountAndSay(self,n):""":type n:int:rtype:str"""ifn==1:return'1'i=1# 从1开始构造 lastStr='1'# 上一个字符串随着迭代次数更新whilen>i:temp=''# 临时子串 count=0# 统计相同字符的个数forjinrange(len(lastStr)):# 找出上一个字符串...
Counting several repeated objects at once is a common problem in programming. Python offers a bunch of tools and techniques you can use to approach this problem. However, Python’s Counter from collections provides a clean, efficient, and Pythonic solution....
If there's a way to trigger this with assigning to function attributes, then I agree this is a problem and should be fixed. You can easily replace the co_consts line with co_consts = outer.__code__.co_consts[:-1] + ([None, None],) and trigger another assert: python: Python/...
original result//& print for current levelresult=temp; cout<<result<<endl;//clear the temporary resulttemp=""; } }intmain() {intn; cout<<"count and Say problem...\n"; cout<<"enter n, no of rows\n"; cin>>n;//function to print count and say sequencecountAndSay(n);return0; ...