To generate then term, justcount and saythen-1 term. 总之就是第n条结果依赖于第n-1条结果。例如: 结果3=21,结果4=1211,从结果3读出1个2和1个1,所以结果4=1211。 话不多说直接上代码吧: 解题代码: classSolution{public:stringcountAndSay(intn){intcount,getnum; string res="1";if(n==1)retu...
string-operation. The only trick thing is Line11. seq[seq.size()] always '\0'. It will help to save an "if" statement. [Code] 1: string countAndSay(int n) {2: // Start typing your C/C++ solution below3: // DO NOT write int main() function4: string seq = "1";5: int ...
首先这个题目肯定要用递归来实现,所以: 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))r...
countAndSay(1) = "1" countAndSay(n) 是对 countAndSay(n-1) 的描述,然后转换成另一个数字字符串。 这个意思就是想求 countAndSay(n) 的话,必须先求 countAndSay(n-1),这就是标准的递归。 使用递归求解,一定不要用...
public class Solution { public String countAndSay(int n) { String res=1+""; if(n==1) return res; else for(int i=1;i<n;i++) res=say(res); return res; } public String say(String res) { String newRes=""; int i=0;
1. Description Count and Say 2. Solution class Solution{public:stringcountAndSay(intn){if(n==1){return"1";}string result;string current="1";for(inti=1;i<n;i++){result=statistic(current);current=result;}returnresult;}private:stringstatistic(string s){string result;intcount=1;charcurrent...
#include<iostream>#include<string>usingnamespacestd;classSolution{public:stringcountAndSay(intn){string res="1";string res2="1";//从2开始数for(inti=2;i<=n;i++){res2="";chara=res[0];//数 第一个数intac=1;//计数intj=1;//while(res[j]!='\0')//结束{if(res[j]==a)//计数...
public class Solution { public String countAndSay(int n) { if(n == 0){ return null; } String read = "1"; for(int i = 1; i < n; i++){ read = process(read); } return read; } public String process(String str){ String result = ""; char temp = str.charAt(0); int coun...
A set of practice note, solution, complexity analysis and test bench to leetcode problem set - leetcode/CountSubArrayFixBound.drawio at b58bcceb0ea27d0756ad72fb6a64b3b547fae221 · brianchiang-tw/leetcode
class Solution { public: int countPrimes(int n) { vector<bool> mp(n, 0); int res = 0; for(int i = 2 ; i < n triplebee 2018/01/12 4130 LeetCode笔记:204. Count Primes javanumbers 我们知道最简单的质数就是2,3,5。。。那怎么计算往后的质数呢?质数的定义是除了自己以外没有任何因子,...