题解 | #验证IP地址#

单纯的手撕字符串，担心嵌入式可能会考。记录一下。
IPv6给的例子02001，多余0不对，我的理解就是超过4个就是不对了。

string solve(string IP) {

    // write code here
    char * rtn1 = "IPv4";
    char * rtn2 = "IPv6";
    char * rtn3 = "Neither";
    string rtn = rtn3;
    int length = IP.size();
    //Part1,判断IPv4 还是v6
    bool flag = 0; //0 : v4  1: v6
    for(int i=0;i<length;i++)
    {
        if( IP[i] == '.')
        {
           flag = 0;
           break   ;
        }
        else if(IP[i] == ':')
        {
           flag = 1;
           break   ;
        }
        else
            continue;
    }
    //part2,
    if(flag == 0) //判断IPv4
    {  
       int i=0;
       int j=0;
       int num = 0;
        while(IP[i] != '\0')
       {
          while(IP[i] != '\0' && IP[i] != '.')
          i++;
          //判断有是不是0      //判断 ..这种情况  
          if(IP[j] == '0' || j == i)
          return rtn;
          //判断0-255
          int temp = 0;int k = j;
          while(k<i)
          {
              temp = temp * 10 + (IP[k]-'0');
              k++;
          }
          if(temp<0 || temp >255)
              return rtn;

          if(IP[i] != '\0')  
          {
              i++;
              j=i;
          }
          num++;

       }
       if(num == 4)  //判断是不是 x.x.x.x格式
       {
            string rtn = rtn1;
            return rtn;
       }
       else
       {
            return rtn;
       }

    }else{  //判断IPv6

        int i=0;
        int j=0;
        int num = 0;
        while(IP[i] != '\0')
        {
          while(IP[i] != '\0' && IP[i] != ':')
          i++;
          //判断有是不是多余4个 //判断::这种情况
          if(i-j > 4 || i == j)
          return rtn;

          if(IP[i] != '\0') //到了末尾就不会继续向下移动了
          {
              i++;
              j=i;
          }
          num++;

       }
       if(num == 8)  //判断是不是 x:x:x:x:x:x:x:x 格式
       {
            string rtn = rtn2;
            return rtn;
       }
       else
       {
            return rtn;
       }


    }
}