[Leetcode] 811. Subdomain Visit Count 解题报告

题目：

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.

思路：

一道简单的字符串处理和哈希表应用的题目：我们首先定义一个哈希表，记录从各个subDomain到其出现次数的映射。然后对于cpdomains中的每个元素，分别解析出该domain出现的次数，以及它所有的superDomains（我们定义discuss.leetcode.com的superDomains为leetcode.com，以及.com），然后将它本身以及所有对应的superDomains的访问次数都更新。最后返回符合格式的哈希表中的内容即可。

代码：

class Solution {
public:
    vector<string> subdomainVisits(vector<string>& cpdomains) {
        for (auto &s : cpdomains) {
            int index = s.find_first_of(' ');
            int count = stoi(s.substr(0, index));
            string domain = s.substr(index + 1);
            getSubdomainVisits(domain, count);
        }
        vector<string> ret;
        for (auto it = hash.begin(); it != hash.end(); ++it) {
            ret.push_back(to_string(it->second) + " " + it->first);
        }
        return ret;
    }
private:
    void getSubdomainVisits(const string &s, int count) {
        int index = 0;
        do {
            string subdomain = index == 0 ? s.substr(index) : s.substr(index + 1);
            hash[subdomain] += count;
            index = s.find_first_of('.', index + 1);
        } while (index != string::npos);
    }
    unordered_map<string, int> hash;
};