[leetcode] Count and Say

Count and Say

使用模拟法：

#include <iostream>
using namespace std;

class Solution {
public:
    string countAndSay(int n) {
        
        if (n==1) {
            return "1";
        }
        
        string preStr="1";
        string nextStr="";
        
        int count=0;
        
        for (int i=2; i<=n; i++) {
            int j=0;
            while (j<preStr.length()) {//注意此处使用while；因为步长不定，不适用for
                char key=preStr[j];
                while (j<preStr.length()&&key==preStr[j]) {//注意此处使用while；if为条件选择，无循环功能
                    count++;
                    j++;
                }
                //cout<<count<<":"<<key<<endl;
                
                nextStr+=count+'0';//转字符
                nextStr+=key;
                
                //cout<<nextStr<<endl;
                count=0;
            }
            
            preStr=nextStr;
            //cout<<"preStr is "<<preStr<<endl;//注释掉，否则会报Output Limit Exceeded
            
            nextStr="";
            
        }
        
        return preStr;
        
    }
};

int main(int argc, const char * argv[])
{

    Solution so;
    //so.countAndSay(1);
    string res=so.countAndSay(7);
    cout<<"the result is "<<res<<endl;
    return 0;
}