字符串中找出连续最长的数字串

1.我已明白
#include<iostream>
#include<string>
#include<vector>
using namespace std;</vector></string></iostream>

int main()
{
string str;

while(cin>>str)
{
    string tmp;
    vector<string> arr;

    for(int i=0;i<=str.length();i++)
    {
        if(str[i]>='0' && str[i]<='9')
        {
            tmp+=str[i];
        }
        else
        {
            if(tmp.length()!=0)
            {
                arr.push_back(tmp);
            }
            tmp.clear();
        }
    }

    int max=0;

    for(int i=0;i<arr.size();i++)
    {
        if(max<(arr[i]).length())
        {
            max=(arr[i]).length(); 
        }
    }

    for(int i=0;i<arr.size();i++)
    {
        if(max==(arr[i]).length())
        {
            cout<<arr[i];
        }
    }
    cout<<","<<max<<endl;
}
return 0;

}
2.还没懂
链接：https://www.nowcoder.com/questionTerminal/2c81f88ecd5a4cc395b5308a99afbbec
来源：牛客网
第一步: 将所有非数字的字符全部转化为空格
第二步: 使用stringstream将字符串分割成数字组成的字符串, 然后比较长度即可
#include<iostream>
#include<string>
#include<sstream>// sstringstream
using namespace std;
int main(){
string str;
while(cin >> str){
for(auto &c : str){
if(!isdigit(c))
c = ' ';
}
stringstream ss(str);
unsigned int max_sz = 0;
string s, o;
while(ss >> s){
if(max_sz < s.size()){
max_sz = s.size();
o = s;
}
else if(max_sz == s.size())
o += s;
}
if(max_sz != 0)
cout << o << "," << max_sz << endl;
}
return 0;}
3.有疑惑
#include<iostream>
#include<string>
using namespace std;
int main(){
string s;
while(cin >> s){
int max=0;
string temp;
string res;
for(int i=0;i<s.size();i++){
if(s[i] >= '0' && s[i] <= '9'){
temp+=s[i];
while(s[i+1] >= '0' && s[i+1] <= '9'){
i++;
temp+=s[i];
}
if(temp.size()>max){
max=temp.size();
res=temp;
}
else if(temp.size()==max){
res+=temp;
}
}
temp.clear();
}
cout << res << ',' << max <<endl;
}
return 0;
}</string></iostream></sstream></string></iostream>

```