[LeetCode] Subdomain Visit Count

A website domain like "discuss.leetcode.com" consists of various subdomains. At the top level, we have "com", at the next level, we have "leetcode.com", and at the lowest level, "discuss.leetcode.com". When we visit a domain like "discuss.leetcode.com", we will also visit the parent domains "leetcode.com" and "com" implicitly.

Now, call a "count-paired domain" to be a count (representing the number of visits this domain received), followed by a space, followed by the address. An example of a count-paired domain might be "9001 discuss.leetcode.com".

We are given a list cpdomains of count-paired domains. We would like a list of count-paired domains, (in the same format as the input, and in any order), that explicitly counts the number of visits to each subdomain.

Example 1:
Input: 
["9001 discuss.leetcode.com"]
Output: 
["9001 discuss.leetcode.com", "9001 leetcode.com", "9001 com"]
Explanation: 
We only have one website domain: "discuss.leetcode.com". As discussed above, the subdomain "leetcode.com" and "com" will also be visited. So they will all be visited 9001 times.

Example 2:
Input: 
["900 google.mail.com", "50 yahoo.com", "1 intel.mail.com", "5 wiki.org"]
Output: 
["901 mail.com","50 yahoo.com","900 google.mail.com","5 wiki.org","5 org","1 intel.mail.com","951 com"]
Explanation: 
We will visit "google.mail.com" 900 times, "yahoo.com" 50 times, "intel.mail.com" once and "wiki.org" 5 times. For the subdomains, we will visit "mail.com" 900 + 1 = 901 times, "com" 900 + 50 + 1 = 951 times, and "org" 5 times.

Notes:

  • The length of cpdomains will not exceed 100
  • The length of each domain name will not exceed 100.
  • Each address will have either 1 or 2 "." characters.
  • The input count in any count-paired domain will not exceed 10000.
  • The answer output can be returned in any order.
统计子域名的访问次数。
1. 利用stringstream来分割字符串。分割每个字符串中的次数和域名。
2. 利用stl中find_first_of()和string中substr来分割域名得到所需要的子域名。
3. 利用哈希表统计并存储子域名及其出现的次数。
4. 将哈希表中的结果放入返回的数组中。
参考代码如下:
class Solution {
public:
    vector<string> subdomainVisits(vector<string>& cpdomains) {
        vector<string> res;
        if (cpdomains.empty())
            return res;
        unordered_map<string, int> m;
        for (auto& str : cpdomains)
        {
            deal(str, m);
        }
        for (auto& s : m)
        {
            string resStr;
            resStr = to_string(s.second) + " " + s.first;
            res.push_back(resStr);
        }
        return res;
    }
    
    void deal(const string& s, unordered_map<string, int>& m)
    {
        stringstream ss(s);
        int val;
        string str;
        ss >> val >> str;
        vector<string> vec;
        vec.push_back(str);
        while (1)
        {
            int pos = str.find_first_of('.');
            if (pos == -1)
                break;
            string tmp(str.substr(pos+1));
            vec.push_back(tmp);
            str = tmp;
        }
        for (auto& v : vec)
        {
            if (m.count(v) > 0)
                m[v] += val;
            else
                m[v] = val;
        }
    }
};

猜你喜欢

转载自www.cnblogs.com/immjc/p/9148924.html