原文链接: https://leetcode-cn.com/problems/subdomain-visit-count

英文原文

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.

中文题目

一个网站域名，如"discuss.leetcode.com"，包含了多个子域名。作为顶级域名，常用的有"com"，下一级则有"leetcode.com"，最低的一级为"discuss.leetcode.com"。当我们访问域名"discuss.leetcode.com"时，也同时访问了其父域名"leetcode.com"以及顶级域名 "com"。

给定一个带访问次数和域名的组合，要求分别计算每个域名被访问的次数。其格式为访问次数+空格+地址，例如："9001 discuss.leetcode.com"。

接下来会给出一组访问次数和域名组合的列表cpdomains 。要求解析出所有域名的访问次数，输出格式和输入格式相同，不限定先后顺序。

示例 1:
输入: 
["9001 discuss.leetcode.com"]
输出: 
["9001 discuss.leetcode.com", "9001 leetcode.com", "9001 com"]
说明: 
例子中仅包含一个网站域名："discuss.leetcode.com"。按照前文假设，子域名"leetcode.com"和"com"都会被访问，所以它们都被访问了9001次。

示例 2
输入: 
["900 google.mail.com", "50 yahoo.com", "1 intel.mail.com", "5 wiki.org"]
输出: 
["901 mail.com","50 yahoo.com","900 google.mail.com","5 wiki.org","5 org","1 intel.mail.com","951 com"]
说明: 
按照假设，会访问"google.mail.com" 900次，"yahoo.com" 50次，"intel.mail.com" 1次，"wiki.org" 5次。
而对于父域名，会访问"mail.com" 900+1 = 901次，"com" 900 + 50 + 1 = 951次，和 "org" 5 次。

注意事项：

•  cpdomains 的长度小于 100。
• 每个域名的长度小于100。
• 每个域名地址包含一个或两个"."符号。
• 输入中任意一个域名的访问次数都小于10000。

通过代码

官方题解

哈希映射：

对于包含一个 . 的域名 x.y，我们需要统计的是 x.y 和 y；对于包含两个 . 的域名 a.b.c，我们需要统计的是 a.b.c，b.c 和 c。在统计这些字符串时，我们可以使用哈希映射（HashMap）。统计结束之后，我们遍历哈希映射并输出结果。

[sol1]
class Solution {
    public List<String> subdomainVisits(String[] cpdomains) {
        Map<String, Integer> counts = new HashMap();
        for (String domain: cpdomains) {
            String[] cpinfo = domain.split("\\s+");
            String[] frags = cpinfo[1].split("\\.");
            int count = Integer.valueOf(cpinfo[0]);
            String cur = "";
            for (int i = frags.length - 1; i >= 0; --i) {
                cur = frags[i] + (i < frags.length - 1 ? "." : "") + cur;
                counts.put(cur, counts.getOrDefault(cur, 0) + count);
            }
        }

        List<String> ans = new ArrayList();
        for (String dom: counts.keySet())
            ans.add("" + counts.get(dom) + " " + dom);
        return ans;
    }
}

[sol1]
class Solution(object):
    def subdomainVisits(self, cpdomains):
        ans = collections.Counter()
        for domain in cpdomains:
            count, domain = domain.split()
            count = int(count)
            frags = domain.split('.')
            for i in xrange(len(frags)):
                ans[".".join(frags[i:])] += count

        return ["{} {}".format(ct, dom) for dom, ct in ans.items()]

复杂度分析

• 时间复杂度：$O(N)$，其中 $N$ 是数组 cpdomains 的长度，这里假设 cpdomains 中每个元素的长度都是常数级别的。

• 空间复杂度：$O(N)$，用于存储哈希映射。

统计信息

通过次数	提交次数	AC比率
16194	22964	70.5%

提交历史

提交时间	提交结果	执行时间	内存消耗	语言