LeetCode 811. Subdomain Visit Count

原题链接在这里：https://leetcode.com/problems/subdomain-visit-count/

题目：

A website domain "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.

A count-paired domain is a domain that has one of the two formats "rep d1.d2.d3" or "rep d1.d2" where rep is the number of visits to the domain and d1.d2.d3 is the domain itself.

• For example, "9001 discuss.leetcode.com" is a count-paired domain that indicates that discuss.leetcode.com was visited 9001 times.

Given an array of count-paired domains cpdomains, return an array of the count-paired domains of each subdomain in the input. You may return the answer in any order. 

Example 1:

Input: cpdomains = ["9001 discuss.leetcode.com"]
Output: ["9001 leetcode.com","9001 discuss.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: cpdomains = ["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.

Constraints:

• 1 <= cpdomain.length <= 100
• 1 <= cpdomain[i].length <= 100
• cpdomain[i] follows either the "repi d1i.d2i.d3i" format or the "repi d1i.d2i" format.
• repi is an integer in the range [1, 104].
• d1i, d2i, and d3i consist of lowercase English letters.

题解：

Have a map to maintain the count.

For each dpdomain, first get its count and its domain.

For each domain and its subdomains, accumlate the count.

Note: there are multiple strings. Make sure to name them and don't mix them.

Time Complexity: O(m*n^2). m = dpdomains.length. n = string length. for each char in the string, substring also takes O(n).

Space: O(m*n^2). map size.

AC Java:

class Solution {
    public List<String> subdomainVisits(String[] cpdomains) {
        List<String> res = new ArrayList<>();
        if(cpdomains == null || cpdomains.length == 0){
            return res;
        }
        
        Map<String, Integer> hm = new HashMap<>();
        for(String s : cpdomains){
            int ind = s.indexOf(' ');
            int count = Integer.valueOf(s.substring(0, ind));
            String d = s.substring(ind + 1);
            hm.put(d, hm.getOrDefault(d, 0) + count);
            for(int i = 0; i < d.length(); i++){
                if(d.charAt(i) == '.'){
                    String subD = d.substring(i + 1);
                    hm.put(subD, hm.getOrDefault(subD, 0) + count);
                }
            }
        }
        
        for(Map.Entry<String, Integer> entry : hm.entrySet()){
            res.add(entry.getValue() + " " + entry.getKey());
        }
        
        return res;
    }
}