[LeetCode] Subdomain Visit Count 子域名访问量统计

 

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.

 

这道题让咱们统计子域名的访问量，所谓的子域名，就是一个完整的域名以点断开的，每一个断开的地方到末尾之间的子字符串就是一个子域名，如今给了咱们不少完整域名的访问量，让咱们统计全部子域名的访问量，题目中给的例子很好的说明了问题。那么这种统计字符串出现个数的问题，咱们应该不难想到须要用一个HashMap来创建字符串和其出现次数的映射。那么接下来要作的就是将每个全域名提取出来，而后拆分红子域名。提取全域名操做不难，由于给的格式都是同样的，前面是数字，中间一个空格，后面是全域名。咱们只须要找到空格的位置，前面的部分转为整型数cnt，后面的就是全域名了。取出全域名以后就要进行拆分红子域名了，咱们能够进行遍历，每当找到小数点的位置时，将后面的子字符串的映射值增长cnt，以此类推直到拆完全部的子域名。注意以前的全域名的映射值别忘了也要加上cnt，最后的最后咱们只要将HashMap中的映射对组成题目中要求返回的格式便可，参见代码以下：

 

解法一：

class Solution {
public:
    vector<string> subdomainVisits(vector<string>& cpdomains) {
        vector<string> res;
        unordered_map<string, int> subdomainCnt;
        for (string cpdomain : cpdomains) {
            int spaceIdx = cpdomain.find(" ");
            int cnt = stoi(cpdomain.substr(0, spaceIdx));
            string rem = cpdomain.substr(spaceIdx + 1);
            for (int i = 0; i < rem.size(); ++i) {
                if (rem[i] == '.') {
                    subdomainCnt[rem.substr(i + 1)] += cnt;
                }
            }
            subdomainCnt[rem] += cnt;
        }
        for (auto a : subdomainCnt) {
            res.push_back(to_string(a.second) + " " + a.first);
        }
        return res;
    }
};

 

下面这种解法和上面的基本相同，惟一改变的地方就是拆分子域名的时候，没用使用遍历的for循环，而是继续使用了find函数来查找下一个小数点的位置，参见代码以下：

 

解法二：

class Solution {
public:
    vector<string> subdomainVisits(vector<string>& cpdomains) {
        vector<string> res;
        unordered_map<string, int> subdomainCnt;
        for (string cpdomain : cpdomains) {
            int spaceIdx = cpdomain.find(" ");
            int cnt = stoi(cpdomain.substr(0, spaceIdx));
            while (spaceIdx != string::npos) {
                subdomainCnt[cpdomain.substr(spaceIdx + 1)] += cnt;
                spaceIdx = cpdomain.find('.', spaceIdx + 1);
            }
        }
        for (auto a : subdomainCnt) {
            res.push_back(to_string(a.second) + " " + a.first);
        }
        return res;
    }
};

 

参考资料：

https://leetcode.com/problems/subdomain-visit-count/solution/

https://leetcode.com/problems/subdomain-visit-count/discuss/157942/C++-concise-solution

https://leetcode.com/problems/subdomain-visit-count/discuss/121738/C++JavaPython-Easy-Understood-Solution

 

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