1: string countAndSay(int n) {2: // Start typing your C/C++ solution below3: // DO NOT write int main() function4: string seq = "1";5: int it = 1;6: while(it<n)7: {8: stringstream newSeq;9: char last = seq[0];10: int count =0;11: for(int i =0; i<= seq.si...
C++演示样例代码: classSolution{public://依据当前的number计算下一个numberstringcountNext(stringnumber) {string::size_type length =number.length(); size_t count =1;//记录同样数字的个数size_t index =0;//遍历number的伪指针char curnum;//暂时保存当前的数字stringresult;//终于结果//这层循环控制对...
leetcode 38 Count and Say 38 Count and Say 36.20% 外观序列,后一个数字等于前一个数字的元素重复个数+元素本身 正则表达式pattern re模块函数: re.sub()替换,re.match()匹配,re.search()查找,re.group()分组, re.split()切分 lambda(): 创建匿名函数,返回一个lambda对象。lambda......
Count and Say 报数 Leetcode 38. Count and Say 报数 标签: Leetcode 题目地址:https://leetcode-cn.com/problems/count-and-say/ 题目描述 报数序列是一个整数序列,按照其中的整数的顺序进行报数,得到下一个数。其前五项如下: 1 被读作 "one 1" ("......
leetcode 38 Count and Say 38 Count and Say 36.20% 外观序列,后一个数字等于前一个数字的元素重复个数+元素本身 正则表达式pattern re模块函数: re.sub()替换,re.match()匹配,re.search()查找,re.group()分组, re.split()切分 lambda(): 创建匿名函数,返回一个lambda对象。lambda......
112112111112213122111311222111132132113113121113122113211311123113112211111312211331121321132122213113112221232112111312211312113211 参考资料: https://discuss.leetcode.com/topic/20195/c-solution-easy-understand 本文转自博客园Grandyang的博客,原文链接:[LeetCode] Count and Say 计数和读法 ,如需转载请自行联系原博主。
【leetcode】38-Count and Say problem Count and Say xiangjian Look-and-say sequence https:///wiki/Look-and-say_sequence code class Solution { public: string countAndSay(int n) { if(n<1) return "";
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 ...
LeetCode Count and Say 数数字 1classSolution {2public:3stringcountAndSay(intn) {4if(n==1)return"1";5stringstr0="",str1="1";6inti,t,count;7charc='*';8for(i=0;i<n-1;i++){//一共要数n-1次,假如n=2,那么只要数str1这一次就行了9count=1;10if(i%2!=0){//i为奇数,数...
classSolution:defcountAndSay(self,n):""":type n: int:rtype: str"""ifn<=1:return'1'defcount(s,last=''):ifs=='':returnlastc=s[0]forrinrange(1,3):if(r>=len(s))or(s[r]!=c):breakelse:r+=1returncount(s[r:],last+'%d%s'%(r,c))returncount(self.countAndSay(n-1)) ...